Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 32

Question 1:

Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches	58	59	60	61	62	63	64	65	66
No. of students	15	20	32	35	35	22	20	10	8

Answer:

In order to find the mean deviation from the median, we will first have to calculate the median.
M is the value of $x$ corresponding to the cumulative frequency just greater than or equal to $\frac{N}{2}$ .

$x_{i}$	f_i	Cumulative Frequency	$\|d_{i}\| = \|x_{i} - 61\|$	$f_{i} \|d_{i}\|$
58	15	15	3	45
59	20	35	2	40
60	32	67	1	32
61	35	102	0	0
62	35	137	1	35
63	22	159	2	44
64	20	179	3	60
65	10	189	4	40
66	8	197	5	40
	$N = Σ f_{i} = 197$			$\sum_{i = 1}^{n} f_{i} \|d_{i}\| = 336$

Here,

\frac{N}{2} = \frac{197}{2} = 98.5

The cumulative frequency just greater than 98.5 is 102. The corresponding value of x is 61.
Therefore, the median is 61.

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |d_{i}| = \frac{1}{197} \times 336 = 1.7055

Question 2:

The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:

Number of calls	0	1	2	3	4	5	6	7
Frequency	14	21	25	43	51	40	39	12

Compute the mean deviation about median.

Answer:

We will first calculate the median.

$x_{i}$	f_i	Cumulative Frequency	$\|d_{i}\| = \|x_{i} - 4\|$	$f_{i} \|d_{i}\|$
0	14	14	4	56
1	21	35	3	63
2	25	60	2	50
3	43	103	1	43
4	51	154	0	0
5	40	194	1	40
6	39	233	2	78
7	12	245	3	36
	$N = Σ f_{i} = 245$			$\sum_{i = 1}^{n} f_{i} \|d_{i}\| = 366$

Here,

\frac{N}{2} = \frac{245}{2} = 122.5

The cumulative frequency just greater than 122.5 is 154 and the corresponding value of x is 4.

∴

M e d i a n, M = 4

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |d_{i}| = \frac{1}{245} \times 366 = 1.493

Question 3:

Calculate the mean deviation about the median of the following frequency distribution:

x_i	5	7	9	11	13	15	17
f_i	2	4	6	8	10	12	8

Answer:

We will first calculate the median for the data.

$x_{i}$	f_i	Cumulative Frequency	$\|d_{i}\| = \|x_{i} - 13\|$	$f_{i} \|d_{i}\|$
5	2	2	8	16
7	4	6	6	24
9	6	12	4	24
11	8	20	2	16
13	10	30	0	0
15	12	42	2	24
17	8	50	4	32
	$N = Σ f_{i} = 50$			$\sum_{i = 1}^{n} f_{i} \|d_{i}\| = 136$

Here,

\frac{N}{2} = \frac{50}{2} = 25

The cumulative frequency just greater than 25 is 30 and the corresponding value of x is 13.

∴ Median, M = 13

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |d_{i}| = \frac{1}{50} \times 136 = 2.72

Question 4:

Find the mean deviation from the mean for the following data:
(i)

x_i	5	7	9	10	12	15
f_i	8	6	2	2	2	6

(ii)

x_i	5	10	15	20	25
f_i	7	4	6	3	5

(iii)

x_i	10	30	50	70	90
f_i	4	24	28	16	8

(iv)

Size	20	21	22	23	24
Frequency	6	4	5	1	4

[NCERT EXEMPLAR]
(v)

Size	1	3	5	7	9	11	13	15
Frequency	3	3	4	14	7	4	3	4

[NCERT EXEMPLAR]

Answer:

i)

x_i	f_i	f_ix_i	$\|x_{i} - \bar{x}\|$	$f_{i} \|x_{i} - 9\|$
5	8	40	4	32
7	6	42	2	12
9	2	18	0	0
10	2	20	1	2
12	2	24	3	6
15	6	90	6	36
	$N = Σ f_{i} = 26$	$\sum_{i = 1}^{n} f_{i} x_{i} = 234$		$\sum_{i = 1}^{n} f_{i} \|x_{i} - 9\| = 88$

x = \frac{{\sum f_{i}}_{i = 1}^{n} x_{i}}{N} = \frac{234}{26} = 9

M . D . = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - x| = \frac{1}{26} \times 88 = 3.39

ii)

x_i	f_i	f_ix_i	$\|x_{i} - \bar{x}\|$	$f_{i} \|x_{i} - 14\|$
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55
	$N = 25$	$\sum_{i = 1}^{n} f_{i} x_{i} = 350$		$\sum_{i = 1}^{n} f_{i} \|x_{i} - 14\| = 158$

x = \frac{{\sum f_{i}}_{i = 1}^{n} x_{i}}{N} = \frac{350}{25} = 14

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - x| = \frac{1}{25} \times 158 = 6.32

iii)

x_i	f_i	f_ix_i	$\|x_{i} - \bar{x}\|$	$f_{i} \|x_{i} - 50\|$
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
	$N = 80$	$\sum_{i = 1}^{n} f_{i} x_{i} = 4000$		$\sum_{i = 1}^{n} f_{i} \|x_{i} - 50\| = 1280$

x = \frac{{\sum f_{i}}_{i = 1}^{n} x_{i}}{N} = \frac{4000}{80} = 50

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - x| = \frac{1}{80} \times 1280 = 16

(iv)

Size(x_i)	Frequency (f_i)	f_ix_i	$\|x_{i} - \bar{x}\| = \|x_{i} - 21.65\|$	$f_{i} \|x_{i} - x\| = f_{i} \|x_{i} - 21.65\|$
20	6	120	1.65	9.9
21	4	84	0.65	2.6
22	5	110	0.35	1.75
23	1	23	1.35	1.35
24	4	96	2.35	9.4
	$N = 20$	$\sum_{i = 1}^{n} f_{i} x_{i} = 433$		$\sum_{i = 1}^{n} f_{i} \|x_{i} - x\| = 25$

x = \frac{{\sum f_{i}}_{i = 1}^{n} x_{i}}{N} = \frac{433}{20} = 21.65

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - x| = \frac{1}{20} \times 25 = 1.25

(v)

Size(x_i)	Frequency (f_i)	f_ix_i	$\|x_{i} - \bar{x}\| = \|x_{i} - 8\|$	$f_{i} \|x_{i} - x\| = f_{i} \|x_{i} - 8\|$
1	3	3	7	21
3	3	9	5	15
5	4	20	3	12
7	14	98	1	14
9	7	63	1	7
11	4	44	3	12
13	3	39	5	15
15	4	60	7	28
	$N = 42$	$\sum_{i = 1}^{n} f_{i} x_{i} = 336$		$\sum_{i = 1}^{n} f_{i} \|x_{i} - x\| = 124$

x = \frac{{\sum f_{i}}_{i = 1}^{n} x_{i}}{N} = \frac{336}{42} = 8

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - x| = \frac{1}{42} \times 124 = 2.95

Question 5:

Find the mean deviation from the median for the following data:
(i)

x_i	15	21	27	30	35
f_i	3	5	6	7	8

(ii)

x_i	74	89	42	54	91	94	35
f_i	20	12	2	4	5	3	4

Answer:

i)

x_i	f_i	Cumulative Frequency	$\|x_{i} - 30\|$	$f_{i} \|x_{i} - 30\|$
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
	$N = Σ f_{i} = 29$			$\sum_{i = 1}^{n} f_{i} \|x_{i} - 30\| = 148$

Here,

\frac{N}{2} = \frac{29}{2} = 14.5

The cumulative frequency greater than 14.5 is 21 and the corresponding value of x is 30.

M e d i a n, M = 30

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - M| = \frac{1}{29} \times 148 = 5.10

ii)

x_i	f_i	Cumulative Frequency	$\|x_{i} - 74\|$	$f_{i} \|x_{i} - 74\|$
35	4	4	39	156
42	2	6	32	64
54	4	10	20	80
74	20	30	0	0
89	12	42	15	180
91	5	47	17	85
94	3	50	20	60
	$N = Σ f_{i} = 50$			$\sum_{i = 1}^{n} f_{i} \|x_{i} - 74\| = 625$

Here,

\frac{N}{2} = \frac{50}{2} = 25

The cumulative frequency greater than 25 is 30 and the corresponding value of x is 74.

M e d i a n, M = 89

M D = \frac{1}{N} \sum_{i = 1}^{n} f_{i} |x_{i} - M| = \frac{1}{50} \times 775 = 12.5

Question 1:

Compute the mean deviation from the median of the following distribution:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	10	20	5	10

Answer:

Class	Frequency $f_{i}$	Cumulative frequency	Mid-values $x_{i}$	$\|d_{i}\| = \|x_{i} - 25\|$	$f_{i} \|d_{i}\|$
0−10	5	5	5	20	100
10−20	10	15	15	10	100
20−30	20	35	25	0	0
30−40	5	40	35	10	50
40−50	10	50	45	20	200
	$N = {\sum f_{i}}_{i = 1}^{5} = 50$				${\sum f_{i} \|d_{i}\|}_{i = 1}^{5} = 450$

Here, N = 50 \Rightarrow \frac{N}{2} = 25

The cumulative frequency greater than

\frac{N}{2} = 25

is 35 and the corresponding class is 20−30.
Therefore, the median class is 20−30.

$∴ l = 20, f = 20, F = 15, N = 50, h = 10 ∴ Median = l + \frac{(\frac{N}{2} - F)}{f} \times h = 20 + \frac{(\frac{50}{2} - 15)}{20} \times 10 = 20 + \frac{(25 - 15)}{20} \times 10 = 25$

Mean deviation from the median = \frac{\sum_{i = 1}^{5} f_{i} |d_{i}|}{N} = \frac{450}{50} = 9

Question 2:

Find the mean deviation from the mean for the following data:
(i)

Classes	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Frequencies	4	8	9	10	7	5	4	3

(ii)

Classes	95-105	105-115	115-125	125-135	135-145	145-155
Frequencies	9	13	16	26	30	12

(iii)

Classes	0-10	10-20	20-30	30-40	40-50	50-60
Frequencies	6	8	14	16	4	2

Answer:

(i) We will compute the mean deviation from the mean in the following way:

Classes	$f_{i}$	Midpoints $x_{i}$	$f_{i} x_{i}$	$\|x_{i} - X\|$ = $\|x_{i} - 358\|$	$f_{i} \|x_{i} - X\|$
0−100	4	50	200	308	1232
100−200	8	150	1200	208	1664
200−300	9	250	2250	108	972
300−400	10	350	3500	8	80
400−500	7	450	3150	92	644
500−600	5	550	2750	192	960
600−700	4	650	2600	292	1168
700−800	3	750	2250	392	1176
	$\sum_{i = 1}^{8} f_{i} = 50$		$\sum_{i = 1}^{8} f_{i} x_{i} = 17900$		$\sum_{i = 1}^{8} f_{i} \|x_{i} - X\| = 7896$

N = \sum_{i = 1}^{6} f_{i} = 50

and

\sum_{i = 1}^{6} f_{i} x_{i} = 17900

X = \frac{\sum_{i = 1}^{8} f_{i} x_{i}}{{\sum f_{i}}_{i = 1}^{8}} = \frac{17900}{50} = 358

∴ Mean deviation = \frac{1}{N} \sum_{i = 1}^{8} f_{i} |x_{i} - X| = \frac{7896}{50} = 157.92

(ii) We will compute the mean deviation from the mean in the following way:

Classes	Frequency $f_{i}$	Midpoints $x_{i}$	$f_{i} x_{i}$	$\|x_{i} - X\|$ = $\|x_{i} - 128.58\|$	$f_{i} \|x_{i} - X\|$
95−105	9	100	900	28.58	257.22
105−115	13	110	1430	18.58	241.54
115−125	16	120	1920	8.58	137.28
125−135	26	130	3380	1.42	36.92
135−145	30	140	4200	11.42	342.6
145−155	12	150	1800	21.42	257.04
	$\sum_{i = 1}^{6} f_{i} = 106$		$\sum_{i = 1}^{6} f_{i} x_{i} = 13630$		$\sum_{i = 1}^{8} f_{i} \|x_{i} - X\| = 1272.6$

N = \sum_{i = 1}^{6} f_{i} = 106

and

\sum_{i = 1}^{6} f_{i} x_{i} = 13630

∴ X = \frac{\sum_{i = 1}^{6} f_{i} x_{i}}{\sum_{i = 1}^{6} f_{i}} = \frac{13630}{106} = 128.58

∴ Mean deviation = \frac{1}{N} \sum_{i = 1}^{8} f_{i} |x_{i} - X| = \frac{1272.6}{106} = 12.005

(iii) We will compute the mean deviation from the mean in the following way:

Classes	Frequency $f_{i}$	Midpoints $x_{i}$	$f_{i} x_{i}$	$\|x_{i} - X\|$ = $\|x_{i} - 358\|$	$f_{i} \|x_{i} - X\|$
0−10	6	5	30	22	132
10−20	8	15	120	12	96
20−30	14	25	350	2	28
30−40	16	35	560	8	128
40−50	4	45	180	18	72
50−60	2	55	110	28	56
	$\sum_{i = 1}^{6} f_{i} = 50$		$\sum_{i = 1}^{6} f_{i} x_{i} = 1350$		$\sum_{i = 1}^{8} f_{i} \|x_{i} - X\| = 512$

N = \sum_{i = 1}^{6} f_{i} = 50

and

\sum_{i = 1}^{6} f_{i} x_{i} = 1350

X = \frac{\sum_{i = 1}^{6} f_{i} x_{i}}{\sum_{i = 1}^{6} f_{i}} = \frac{1350}{50} = 27

∴ Mean deviation = \frac{1}{N} \sum_{i = 1}^{8} f_{i} |x_{i} - X| = \frac{512}{50} = 10.24

Question 3:

Compute mean deviation from mean of the following distribution:

Mark	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	8	10	15	25	20	18	9	5

Answer:

Computation of mean deviation from the mean:

Marks	Number of Students $f_{i}$	Midpoints $x_{i}$	$f_{i} x_{i}$	$\|x_{i} - X\|$ = $\|x_{i} - 49\|$	$f_{i} \|x_{i} - X\|$
10−20	8	15	120	34	272
20−30	10	25	250	24	240
30−40	15	35	525	14	210
40−50	25	45	1125	4	100
50−60	20	55	1100	6	120
60−70	18	65	1170	16	288
70−80	9	75	675	26	234
80−90	5	85	425	36	180
	$N = \sum_{i = 1}^{8} f_{i} = 110$		$\sum_{i = 1}^{8} f_{i} x_{i} = 5390$		$\sum_{i = 1}^{8} f_{i} \|x_{i} - X\| = 1644$

N = \sum_{i = 1}^{8} f_{i} = 110

and

\sum_{i = 1}^{8} f_{i} x_{i} = 5390

X = \frac{\sum_{i = 1}^{8} f_{i} x_{i}}{N} = \frac{5390}{110} = 49

Mean deviation = \frac{\sum_{i = 1}^{8} f_{i} |x_{i} - X|}{N} = \frac{1644}{110} = 14.945 \approx 14.95

Question 4:

The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birth day)	17-19.5	20-25.5	26-35.5	36-40.5	41-50.5	51-55.5	56-60.5	61-70.5
No. of persons	5	16	12	26	14	12	6	5

Calculate the mean deviation from the median age.

Answer:

To make this function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

Age	Number of People $f_{i}$	Cumulative Frequency	Midpoints $x_{i}$	$\|d_{i}\| = \|x_{i} - 38.63\|$	$f_{i} \|d_{i}\|$
16.75−19.75	5	5	18.25	20.38	101.9
19.75−25.75	16	21	22.75	15.88	254.08
25.75−35.75	12	33	30.75	7.88	94.56
35.75−40.75	26	59	38.25	0.38	9.38
40.75−50.75	14	73	45.75	7.12	99.68
50.75−55.75	12	85	53.25	14.62	175.44
55.75−60.75	6	91	58.25	19.62	117.72
60.75−70.75	5	96	65.75	27.12	135.6
	$N = {\sum f_{i}}_{i = 1}^{8} = 96$				$\sum_{i = 1}^{8} f_{i} \|d_{i}\| = 988.36$

N = 96 ∴ \frac{N}{2} = 48 The cumulative frequency just greater than \frac{N}{2} = 38 is 59 and the corresponding class is 35.75 - 40.75 . Thus, the median class is 35.75 - 40.75 l = 35.75, f = 26, F = 33, h = 5 Median = l + \frac{\frac{N}{2} - F}{f} \times h = 35.75 + \frac{(48 - 33)}{26} \times 5 = 35.75 + 2.88 = 38.63 Mean deviation from the median = \frac{\sum_{i = 1}^{8} f_{i} |d_{i}|}{N} = \frac{988.36}{96} = 10.29

Question 5:

Find the mean deviation from the mean and from median of the following distribution:

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

Answer:

Computation of mean distribution from the median:

Marks	Number of Students $f_{i}$	Cumulative Frequency	Midpoints $x_{i}$	$\|d_{i}\| = \|x_{i} - 28\|$	$f_{i} \|d_{i}\|$	$f_{i} x_{i}$	$\|x_{i} - 27\|$	$f_{i} \|x_{i} - 27\|$
0−10	5	5	5	23	115	25	22	110
10−20	8	13	15	13	104	120	12	96
20−30	15	28	25	3	45	375	2	30
30−40	16	44	35	7	112	560	8	128
40−50	6	50	45	17	102	270	18	108
	$N = 50$				$\sum_{i = 1}^{5} f_{i} \|d_{i}\| = 478$	1350		$\sum_{i = 1}^{5} f_{i} \|x_{i} - 27\| = 472$

N = 50, \frac{N}{2} = 25

The cumulative frequency just greater than

\frac{N}{2} = 25

is 28 and the corresponding class is 20−30.
Thus, the median class is 20−30.

Using formula:

$∴ l = 20, F = 13, f = 15, h = 10 Median = l + \frac{\frac{N}{2} - F}{f} \times h Substituting the values : Median = 20 + \frac{25 - 13}{15} \times 10 = 20 + 8 = 28$

Mean distribution from the median = \frac{\sum_{i = 1}^{5} f_{i} |d_{i}|}{N} = \frac{478}{50} = 9.56

Mean (X) = \frac{\sum_{i = 1}^{5} f_{i} x_{i}}{N} = \frac{1350}{50} = 27 Mean deviation from the mean = \frac{1}{50} \times \sum_{i = 1}^{5} f_{i} |x_{i} - 27| = \frac{472}{50} = 9.44

Mean deviation from the median and the mean are 9.56 and 9.44, respectively.

Question 6:

Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age:	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
Number of persons	5	6	12	14	26	12	16	9

Answer:

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains same, while the function becomes continuous.

Thus, the mean distribution table will be as follows:

Age	Number of Persons $f_{i}$	Midpoint $x_{i}$	Cumulative Frequency	_{$\|d_{i}\| = \|x_{i} - 38\|$}	$f_{i} d_{i}$
15.5−20.5	5	18	5	20	100
20.5−25.5	6	23	11	15	90
25.5−30.5	12	28	23	10	120
30.5−35.5	14	33	37	5	70
35.5−40.5	26	38	63	0	0
40.5−45.5	12	43	75	5	60
45.5−50.5	16	48	91	10	160
50.5−55.5	9	53	100	15	135
	$N = \sum_{i = 1}^{8} f_{i} = 100$				$\sum_{i = 1}^{8} f_{i} d_{i} = 735$

N = 100 \Rightarrow \frac{N}{2} = 50

Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5−40.5.

$∴ l = 35.5, F = 37, f = 26, h = 5 Median = l + \frac{\frac{N}{2} - F}{f} \times h = 35.5 + \frac{(50 - 37)}{26} \times 5 = 35.5 + 2.5 = 38 Mean deviation about the median age = \frac{\sum_{i = 1}^{8} f_{i} |d_{i}|}{N} = \frac{735}{100} = 7.35$

Thus, the mean deviation from the median age is 7.35 years.

Question 7:

Calculate the mean deviation about the mean for the following frequency distribution:

Class interval:	0–4	4–8	8–12	12–16	16–20
Frequency	4	6	8	5	2

Answer:

Let the assumed mean A = 10 and h = 4.

Class Interval	Mid-Value(x_i)	Frequency(f_i)	$d_{i} = \frac{x_{i} - 10}{4}$	$f_{i} d_{i}$	$\|x_{i} - X\| = \|x_{i} - 9.2\|$	$f_{i} \|x_{i} - X\|$
0–4	2	4	−2	−8	7.2	28.8
4–8	6	6	−1	−6	3.2	19.2
8–12	10	8	0	0	0.8	6.4
12–16	14	5	1	5	4.8	24
16–20	18	2	2	4	8.8	17.6
		N = 25		$\sum_{} f_{i} d_{i}$ = −5		$\sum_{} f_{i} \|x_{i} - X\| =$ 96

Here, N = 25 and

\sum_{} f_{i} d_{i}

= −5

Mean,

X = A + h (\frac{1}{N} \sum_{} f_{i} d_{i}) = 10 + 4 (\frac{1}{25} \times (- 5)) = 10 - \frac{20}{25} = 10 - 0.8 = 9.2

∴ Mean deviation about mean

= \frac{1}{N} \sum_{} f_{i} |x_{i} - X| = \frac{1}{25} \times 96 = 3.84

Question 8:

Calculate mean deviation from the median of the following data: [NCERT EXEMPLAR]

Class interval:	0–6	6–12	12–18	18–24	24–30
Frequency	4	5	3	6	2

Answer:

Calculation of mean deviation about the median.

Class Interval	Mid-Values (x_i)	Frequency (f_i)	Cummulative Frequency (c.f.)	$\|x_{i} - 14\|$	$f_{i} \|x_{i} - 14\|$
0–6	3	4	4	11	44
6–12	9	5	9	5	25
12–18	15	3	12	1	3
18–24	21	6	18	7	42
24–30	27	2	20	13	26
		N = 20			$\sum_{} f_{i} \|x_{i} - 14\| = 140$

Here, N = 20. So,

\frac{N}{2} = 10

The cummulative frequency just greater than

\frac{N}{2}

is 12. Thus, 12–18 is the median class.

Now, l = 12, h = 6, f = 3 and F = 9

∴ Median = l + \frac{\frac{N}{2} - F}{f} \times h = 12 + (\frac{10 - 9}{3}) \times 6 = 14

Now,

Mean deviation about median =

\frac{1}{N} \sum_{} f_{i} |x_{i} - 14| = \frac{1}{20} \times 140 = 7

Question 1:

Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17.
(ii) 6, 7, 10, 12, 13, 4, 8, 12.
(iii) 227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
(iv) 15, 22, 27, 11, 9, 21, 14, 9.

Answer:

(i) 2,4,5,6,8,17

$Mean = X = \frac{2 + 4 + 5 + 6 + 8 + 17}{6} = \frac{42}{6} = 7$

$x_{i}$	$(x_{i} - X) = (x_{i} - 7)$	${(x_{i} - 7)}^{2}$
2	-5	25
4	-3	9
5	-2	4
6	-1	1
8	1	1
17	10	100
		$\sum_{i = 1}^{6} {(x_{i} - X)}^{2} = 140$

n = 6

Variance (X) = \frac{\sum_{i = 1}^{6} {(x_{i} - X)}^{2}}{n} = \frac{140}{6} = 23.33

Standard deviation = \sqrt{Variance (X}) = \sqrt{23.33} = 4.83

(ii) 6,7,10,12,13,4,8,12

Mean = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9

$x_{i}$	$(x_{i} - X) = (x_{i} - 9)$	${(x_{i} - X)}^{2}$
6	-3	9
7	-2	4
10	1	1
12	3	9
13	4	16
4	$-$ 5	25
8	$-$ 1	1
12	3	9
		$\sum_{i = 1}^{8} {(x_{i} - X)}^{2} = 74$

n = 8

∴ Variance (X) = \frac{\sum_{i = 1}^{8} {(x_{i} - X)}^{2}}{n} = \frac{74}{8} = 9.25 Standard deviation = \sqrt{Variance (X)} = \sqrt{9.25} = 3.04

(iii) 227,235,255,269,292,299,312,321,333,348,

Mean = \frac{227 + 235 + 255 + 269 + 292 + 299 + 312 + 321 + 333 + 348}{10} = \frac{2891}{10} = 289.1

$x_{i}$	$(x_{i} - X) = (x_{i} - 289.1)$	${(x_{i} - X)}^{2}$
227	− 62.1	3856.41
235	− 54.1	2926.81
255	− 34.1	1162.81
269	− 20.1	404.01
292	2.9	8.41
299	9.9	98.01
312	22.9	524.41
321	31.9	1017.61
333	43.9	1927.21
348	58.9	3469.21
		$\sum_{i = 1}^{10} {(x_{i} - \bar{x})}^{2} = 15394.9$

n = 10 ∴ Variance (X) = \frac{\sum_{i = 1}^{10} {(x_{i} - \bar{X})}^{2}}{n} = \frac{15394.9}{10} = 1539.49 Standard deviation = \sqrt{Variance (X)} = \sqrt{1539.49} = 39.24

(iv) 15,22,27,11,9,21,14,9

Mean = \frac{15 + 22 + 27 + 11 + 9 + 21 + 14 + 9}{8} = \frac{128}{8} = 16

$x_{i}$	$(x_{i} - X) = (x_{i} - 16)$	${(x_{i} - \bar{X})}^{2}$
15	−1	1
22	6	36
27	11	121
11	5	25
9	−7	49
21	5	25
14	−2	4
9	−7	49
		$\sum_{i = 1}^{8} {(x_{i} - \bar{X})}^{2} = 310$

∴ n = 8 Variance (X) = \frac{\sum_{i = 1}^{8} {(x_{i} - x)}^{2}}{n} = \frac{310}{8} = 38.75 Standard deviation = \sqrt{Variance (X)} = \sqrt{38.75} = 6.22

Question 2:

The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

Answer:

Let $x_{1}, x_{2}, x_{3}, . . ., x_{20}$ be the 20 given observations.

$Variance (X) = 5$

$Variance (X) = \frac{1}{20} \times \sum {(x_{i} - X)}^{2} = 5 (Here, X is the mean of the given observations .)$

Let u₁_,u₂_,,u₃, ..., u₂₀ be the new observations, such that
$u_{i} = 2 x_{i} (for i = 1, 2, 3, . . ., 20) . . . (1)$

$Mean = \bar{U} = \frac{\sum_{i = 1}^{20} u_{i}}{n} = \frac{\sum_{i = 1}^{20} 2 x_{i}}{20} [substituting u_{i} from eq (1) and taking n as 20] = 2 \times \frac{{\sum x_{i}}_{i = 1}^{20}}{20} = 2 \bar{X}$

$u_{i} - \bar{U} = 2 x_{i} - 2 \bar{X} (for i = 1, 2, . . ., 20) = 2 (x_{i} - \bar{X}) {(u_{i} - \bar{U})}^{2} = {(2 (x_{i} - \bar{X}))}^{2} (squaring both the sides) = 4 {(x_{i} - \bar{X})}^{2} ∴ \sum_{i = 1}^{20} {(u_{i} - \bar{U})}^{2} = {\sum 4}_{i = 1}^{20} {(x_{i} - \bar{X})}^{2} \frac{\sum_{i = 1}^{20} {(u_{i} - \bar{U})}^{2}}{20} = \frac{{\sum 4}_{i = 1}^{20} {(x_{i} - \bar{X})}^{2}}{20} = 4 \frac{\sum_{i = 1}^{20} {(x_{i} - \bar{X})}^{2}}{20} Variance (U) = 4 \times Variance (X) = 4 \times 5 = 20$

Thus, variance of the new observations is 20.

Question 3:

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Answer:

Let x₁_,,x₂_,,x₃, ..., x₁₅ be the given observations.

Variance X is given as 4.
If $\bar{X}$ is the mean of the given observations, then we get:
$\Rightarrow Variance X = \frac{1}{15} \sum_{i = 1}^{15} {(x_{i} - X)}^{2} = 4$

Let u₁_,u₂_,u₃ ... u₁₅ be the new observations such that
$u_{i} = x_{i} + 9 (for i = 1, 2, 3, . . ., 15) . . . . (1) \bar{U} = \frac{1}{n} \sum_{i = 1}^{15} u_{i} = \frac{1}{15} \sum_{i = 1}^{15} (x_{i} + 9) = \frac{1}{15} {\sum x_{i}}_{i = 1}^{15} + \frac{9 \times 15}{15} [as \sum_{i = 1}^{15} 9 = 9 \times 15] = X + 9 . . . (2) u_{i} - \bar{U} = (x_{i} + 9) - (9 + \bar{X}) [from eq (1) and eq (2)] = x_{i} - \bar{X} \Rightarrow \frac{1}{15} \times {(u_{i} - \bar{U})}^{2} = \frac{1}{15} {(x_{i} - \bar{X})}^{2} [squaring both the sides and then dividing by 15] \Rightarrow \frac{1}{15} \times \sum_{i = 1}^{15} {(u_{i} - \bar{U})}^{2} = \frac{1}{15} \times \sum_{i = 1}^{15} {(x_{i} - \bar{X})}^{2} \Rightarrow \frac{1}{15} \times \sum_{i = 1}^{15} {(u_{i} - \bar{U})}^{2} = 4 \Rightarrow Variance (U) = 4$

Thus, variance of the new observation is 4.

Question 4:

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Answer:

Let $x and y$ be the other two observations.

Mean is 4.4.

$∴ \frac{1 + 2 + 6 + x + y}{5} = 4.4$

$\Rightarrow 9 + x + y = 22 \Rightarrow x + y = 13 . . . (1)$

Let Var (X) be the variance of these observations, which is given to be 8.24.

If $\bar{X}$ is the mean, then we have:

$8.24 = \frac{1}{5} (1^{2} + 2^{2} + 6^{2} + x^{2} + y^{2}) - {(x)}^{2} = \frac{1}{5} (1 + 4 + 36 + x^{2} + y^{2}) - {(4.4)}^{2} = \frac{1}{5} (41 + x^{2} + y^{2}) - 19.36 \Rightarrow x^{2} + y^{2} = 97 . . . (2) {(x + y)}^{2} + {(x - y)}^{2} = 2 (x^{2} + y^{2}) \Rightarrow 13^{2} + {(x - y)}^{2} = 2 \times 97 [using eq (1) and eq (2)] \Rightarrow {(x - y)}^{2} = 194 - 169 = 25 \Rightarrow x - y = \pm 5 . . . . (3) Solving eq (1) and eq (3) for x - y = - 5 and x + y = 13 2 x = 18 \Rightarrow x = 9 \Rightarrow y = 4$

Thus, the other two observations are 9 and 4.

Question 5:

The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Answer:

$Mean = \bar{X} = 8 n = 6 σ = S . D = 4 I f x_{1,} x_{2 . . . .} x_{6} are the given observations X = \frac{1}{n} \times \sum_{i = 1}^{6} x_{i} \Rightarrow 8 = \frac{1}{6} \times \sum_{i = 1}^{6} x_{i} Let u_{1,} u_{2} . . . . . u_{6} be the new observations \Rightarrow u_{i} = 3 x_{i} (for i = 1, 2, 3 . . . 6) \Rightarrow Mean of new observations = \bar{U} = \frac{1}{n} \times {\sum u_{i}}_{i = 1}^{6} = \frac{1}{6} \times {\sum 3 x_{i}}_{i = 1}^{6} = 3 \times \frac{1}{6} \times \sum_{i = 1}^{6} x_{i} = 3 \bar{X} = 3 \times 8 = 24$

Thus, mean of the new observations is 24.

$SD = σ_{x} = 4 {σ_{x}}^{2} = Variance X ∴ Variance X = 16 \Rightarrow \frac{1}{6} \sum_{i = 1}^{6} {(x_{i} - \bar{X})}^{2} = 16 . . . (1) Variance (U) = {σ_{u}}^{2} = \frac{1}{6} \sum_{i = 1}^{6} {(u_{i} - \bar{U})}^{2} = \frac{1}{6} \times \sum_{i = 1}^{6} {(3 x_{i} - 3 \bar{X})}^{2} = 3^{2} \times \frac{1}{6} \sum_{i = 1}^{6} {(x_{i} - \bar{X})}^{2} = 9 \times 16 σ_{u} = \sqrt{Variance (U)} = \sqrt{9 \times 16} = 12$

Thus, standard deviation of the new observations is 12.

Question 6:

The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Answer:

Let x and y be the remaining two observations.

$n = 8 Variance = 9.25 \bar{X} = Mean = 9 \Rightarrow \frac{6 + 7 + 10 + 12 + 12 + 13 + x + y}{8} = 9 \Rightarrow 60 + x + y = 72 \Rightarrow x + y = 12 . . . (1)$

$Variance X = \frac{1}{n} \sum_{i = 1}^{8} {x_{i}}^{2} - {(\bar{X})}^{2} \Rightarrow 9.25 = (\frac{1}{8} \times (6^{2} + 7^{2} + 10^{2} + 12^{2} + 12^{2} + 13^{2} + x^{2} + y^{2})) - 9^{2} \Rightarrow 9.25 = \frac{1}{8} (642 + x^{2} + y^{2}) - 81 \Rightarrow 9.25 \times 8 = 642 + x^{2} + y^{2} - 648 \Rightarrow x^{2} + y^{2} = 80 . . . . (2) We know, {(x + y)}^{2} + {(x - y)}^{2} = 2 (x^{2} + y^{2}) \Rightarrow 12^{2} + {(x - y)}^{2} = 2 \times 80 [using equations (1) and (2)] \Rightarrow 144 + {(x - y)}^{2} = 160 \Rightarrow {(x - y)}^{2} = 16 \Rightarrow x - y = \pm 4 If x - y = 4, then x + y = 12 and x - y = 4 give x = 8 and y = 4 If x - y = - 4, then x + y = 12 and x - y = 4 give x = 4 and y = 8$

Thus, the remaining two observations are 8 and 4.

Question 7:

For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

Answer:

We have:
$n = 200, \bar{X} = 40, σ = 15$

$\frac{1}{n} \sum x_{i} = \bar{X} ∴ \frac{1}{200} \sum_{} x_{i} = 40 \Rightarrow \sum_{} x_{i} = 40 \times 200 = 8000 Since the score was misread, this sum is incorrect . \Rightarrow Corrected \sum_{}^{} x_{i} = 8000 - 34 - 53 + 43 + 35 = 8000 - 7 = 7993 ∴ Corrected mean = \frac{Corrected \sum_{}^{} x_{i}}{200} = \frac{7993}{200} = 39.955$

$SD = σ = 15 \Rightarrow Variance = 15^{2} = 225 According to the formula, Variance = (\frac{1}{n} \sum_{} {x_{i}}^{2}) - {(\frac{1}{n} \sum_{} x_{i})}^{2} ∴ \frac{1}{200} \sum_{} {x_{i}}^{2} - {(40)}^{2} = 225 \Rightarrow \frac{1}{200} \sum_{} {(x_{i})}^{2} - 1600 = 225 \underset{}{\Rightarrow \sum} {(x_{i})}^{2} = 200 \times 1825 = 365000 This is an incorrect reading . ∴ Corrected \sum_{} {(x_{i})}^{2} = 365000 - 34^{2} - 53^{2} + 43^{2} + 35^{2} = 365000 - 1156 - 2809 + 1849 + 1225 = 364109$

$Corrected variance = (\frac{1}{n} \times Corrected \sum_{} x_{i}) - {(Corrected mean)}^{2} = (\frac{1}{200} \times 364109) - {(39.955)}^{2} = 1820.545 - 1596.402 = 224.14 Corrected SD = \sqrt{Corrected variance} = \sqrt{224.14} = 14.97$

Question 8:

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Answer:

$n = 100 Mean = \bar{X} = 40 σ = SD = 5.1 \frac{1}{n} \sum x_{i} = \bar{X} \Rightarrow \sum x_{i} = 100 \times 40 = 4000 (This is an incorrect reading due to misread values .) Corrected sum, \sum x_{i} = 4000 - 50 + 40 = 3990 \Rightarrow Corrected mean = \frac{Corrected sum}{100} = \frac{3990}{100} = 39.9 . . . (1)$

To find the corrected SD:

$\sqrt{Variance} = σ \Rightarrow σ^{2} = {(5.1)}^{2} = Variance According to the formula, \frac{1}{n} \sum_{} {x_{i}}^{2} - {(\bar{X})}^{2} = Variance \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} - {(40)}^{2} = 26.01 \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} - 1600 = 26.01 \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} = 1626.01 \Rightarrow \sum_{} {x_{i}}^{2} = 162601 (But, this is incorrect due t o misread values) \Rightarrow Corrected \sum_{} {x_{i}}^{2} = 162601 - 50^{2} + 40^{2} = 161701 . . . . (2) Corrected variance = \frac{1}{100} Corrected \sum_{} {x_{i}}^{2} - {(Corrected mean)}^{2} = \frac{161701}{100} - {(39.9)}^{2} [using equations (1) and (2)] = 1617.01 - 1592 . 01 = 25 Corrected SD = \sqrt{Corrected variance} = \sqrt{25} = 5$

Corrected mean = 39.9
Corrected standard deviation = 5

Question 9:

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.

Answer:

$n = 20 Mean = \bar{X} SD = σ = 2 \frac{1}{n} \underset{}{\sum x_{i} =} \bar{X} ∴ \frac{1}{20} \underset{}{\sum x_{i} = 10} \Rightarrow \sum x_{i} = 200 [This is incorrect due to misread values .] . . . (1) \Rightarrow Variance = σ^{2} = 4 \frac{1}{n} \sum_{} {x_{i}}^{2} - (\bar{X})^{2} = 4 \Rightarrow \frac{1}{20} \sum_{} {x_{i}}^{2} - 10^{2} = 4 \Rightarrow \frac{1}{20} \sum_{} {x_{i}}^{2} = 104 \Rightarrow \sum_{} {x_{i}}^{2} = 104 \times 20 = 2080 [This is incorrect due to misread values .] . . . (2)$

(i) If observation 8 is omitted, then total 19 observations are left.

         Incorrected $\sum_{} x_{i} = 200$

        $Corrected \sum_{} x_{i} + 8 = 200$

$Corrected \sum_{}^{} x_{i} = 192 \Rightarrow (Corrected mean) = \frac{Corrected \sum_{}^{} x_{i}}{19} = \frac{192}{19} = 10.10 Using equation (2), we get : Corrected \sum_{}^{} {x_{i}}^{2} + 8^{2} = 2080 \Rightarrow Corrected \sum_{}^{} {x_{i}}^{2} = 2080 - 64 = 2016 ∴ \frac{1}{19} Corrected \sum_{}^{} {x_{i}}^{2} - {(Corrected mean)}^{2} = Corrected variance \Rightarrow Corrected variance = \frac{1}{19} \times 2016 - {(\frac{192}{19})}^{2} \Rightarrow Corrected variance = \frac{(2016 \times 19) - {(192)}^{2}}{19^{2}} \Rightarrow Corrected variance = \frac{38304 - 36864}{19^{2}} \Rightarrow Corrected variance = \frac{1440}{19^{2}} Corrected SD = \sqrt{Corrected variance} = \sqrt{\frac{1440}{19^{2}}} = \frac{12 \sqrt{10}}{19} = 1.997$

Thus, if 8 is omitted, then the mean is 10.10 and SD is 1.997.

(ii) When incorrect observation 8 is replaced by 12:


$From eq u a t i o n (1) : Incorrected \sum_{}^{} x_{i} = 200 Corrected \sum_{}^{} x_{i} = 200 - 8 + 12 = 204 Corrected \bar{X} = \frac{204}{20} = 10.2 Incorrected \sum_{}^{} {x_{i}}^{2} = 2080 [from (2)] Corrected \sum_{}^{} {x_{i}}^{2} = 2080 - 8^{2} + 12^{2} = 2160 Corrected variance = \frac{1}{20} \times Corrected \sum_{}^{} {x_{i}}^{2} - {(Corrected \bar{X})}^{2} = \frac{1}{20} \times 2160 - {(\frac{204}{20})}^{2} = \frac{(2160 \times 20) - {(204)}^{2}}{20^{2}} = \frac{43200 - 41616}{400} = \frac{1584}{400} Corrected SD = \sqrt{Corrected variance} = \sqrt{\frac{1584}{400}} = \frac{\sqrt{396}}{10} = \frac{19.899}{10} = 1.9899$

If 8 is replaced by 12, then the mean is 10.2 and SD is 1.9899.

Question 10:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.

Answer:

$n = 100 Mean = \bar{X} = 20 SD = σ = 3 M isread values are 21, 21 and 18 . \underset{}{\frac{1}{n} \sum} x_{i} = \bar{X} \Rightarrow \frac{1}{100} \underset{}{\sum x_{i} = 20} \Rightarrow \underset{}{\sum x_{i} = 20 \times 100 = 2000} [This sum is incorrect due to misread values .] . . . . (1) If three misread values are to be omitted, the total number of enteries will be 97 . Also, \sum x_{i} = 2000 - (21 - 21 - 18) = 1940 Corrected \bar{X} = \frac{1940}{97} = 20 . . . . (2)$

$σ = 3 \Rightarrow Variance = σ^{2} = 9 = Variance = \frac{1}{n} \sum_{} {x_{i}}^{2} - {(\bar{X})}^{2} \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} - 20^{2} = 9 \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} = 9 + 400 \Rightarrow \frac{1}{100} \sum_{} {x_{i}}^{2} = 409 \Rightarrow \sum_{} {x_{i}}^{2} = 409 \times 100 = 40900 (This is an incorrect sum due to misread values .) . . . (2) Corrected \sum_{} {x_{i}}^{2} = 40900 - (21^{2} + 21^{2} + 18^{2}) = 40900 - 441 - 441 - 324 = 39694 . . . . (3) From equations (2) and (3), we get : Corrected variance = \frac{1}{n} \sum_{} {x_{i}}^{2} - {(\bar{X})}^{2} = \frac{1}{97} \times 39694 - {(20)}^{2} = 409.216 - 400 = 9.216 Corrected SD = \sqrt{Corrected variance} = \sqrt{9.216} = 3.0357$

Thus, after omitting three values, the mean would be 20 and SD would be 3.0357.

Question 11:

Show that the two formulae for the standard deviation of ungrouped data

$σ = \sqrt{\frac{1}{n} \underset{}{\sum {(x_{i} - X)}^{2}}}$ and $σ' = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2} - X^{2}}}$ are equivalent, where $X = \frac{1}{n} \sum_{} x_{i}$

Answer:

$σ = \sqrt{\frac{1}{n} \underset{}{\sum {(x_{i} - X)}^{2}}} = \sqrt{\frac{1}{n} \underset{}{\sum (x_{i}^{2} - 2 x_{i} X + X^{2})}} = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2}} - \frac{1}{n} \sum 2 x_{i} X + \frac{1}{n} \sum X^{2}} = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2}} - \frac{1}{n} \times 2 X \sum x_{i} + \frac{1}{n} \times X^{2} \sum 1} = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2}} - \frac{1}{n} \times 2 X \times n X + \frac{1}{n} \times X^{2} \times n} (X = \frac{1}{n} \underset{}{\sum x_{i}})$
$= \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2} - 2 X^{2} +} X^{2}} = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2} - X^{2}}} = σ'$

Hence, the formulae $σ = \sqrt{\frac{1}{n} \underset{}{\sum {(x_{i} - X)}^{2}}}$ and $σ' = \sqrt{\frac{1}{n} \underset{}{\sum x_{i}^{2} - X^{2}}}$ are equivalent, where $X = \frac{1}{n} \sum_{} x_{i}$ .

Question 1:

Find the standard deviation for the following distribution:

x :	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f :	1	5	12	22	17	9	4

Answer:

x:	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f:	1	5	12	22	17	9	4

Median value of

x

is 34.5.

$x_{i}$	$f_{i}$	$d_{i} = x_{i} - 34.5$	$u_{i} = \frac{x_{i} - 34.5}{10}$	$f_{i} u_{i_{}}$	${u_{i}}^{2}$	$f_{i} {u_{i}}^{2}$
4.5	1	$-$ 30	$-$ 3	$-$ 3	9	9
14.5	5	$-$ 20	$-$ 2	$-$ 10	4	20
24.5	12	$-$ 10	$-$ 1	$-$ 12	1	12
34.5	22	0	0	0	0	0
44.5	17	10	1	17	1	17
54.5	9	20	2	18	4	36
64.5	4	30	3	12	9	36
	$N = \sum f_{i} = 70$			$\sum f_{i} u_{i} = 22$		$\sum f_{i} {u_{i}}^{2} = 130$

Var (X) = h^{2} [\frac{1}{N} \sum_{i = 1}^{n} f_{i} {u_{i}}^{2} - {(\frac{1}{N} \sum_{i = 1}^{n} f_{i} u_{i})}^{2}]

We have

N = 70, \sum_{i = 1}^{n} f_{i} u_{i} = 22, \sum_{i = 1}^{n} f_{i} {u_{i}}^{2} = 130, h = 10

Plugging all the values in the formula of variance:

Var (X) = 10^{2} [\frac{1}{70} \times 130 - {(\frac{1}{70} \times 22)}^{2}] = 100 [\frac{130}{70} - {(\frac{22}{70})}^{2}] = 100 [\frac{13}{7} - \frac{121}{1225}] = 100 [1.857 - 0.0987] = 100 [1.7583] = 175.83

Standard deviation,

SD = \sqrt{Var (X)}

SD = \sqrt{Var (X)} = \sqrt{175.83} = 13.26

Question 2:

Table below shows the frequency f with which 'x' alpha particles were radiated from a diskette

x :	0	1	2	3	4	5	6	7	8	9	10	11	12
f :	51	203	383	525	532	408	273	139	43	27	10	4	2

Calculate the mean and variance.

Answer:

Mean, $\bar{X} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{10078}{2600} = 3.88$

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$x_{i} - \bar{X}$	${(x_{i} - \bar{X})}^{2}$	$f_{i} {(x_{i} - \bar{X})}^{2}$
0	51	0	−3.88	15.05	767.55
1	203	203	−2.88	8.29	1682.87
2	383	766	−1.88	3.53	1351.99
3	525	1575	−0.88	0.77	404.25
4	532	2128	0.12	0.014	7.448
5	408	2040	1.12	1.25	510
6	273	1638	2.12	4.49	1225.77
7	139	973	3.12	9.73	1352.47
8	43	344	4.12	16.97	729.71
9	27	243	5.12	26.21	707.67
10	10	100	6.12	37.45	374.5
11	4	44	7.12	50.69	202.76
12	2	24	8.12	65.93	131.86
	$\sum f_{i} = N = 2600$	$\sum f_{i} x_{i} = 10078$			$\sum f_{i} {(x_{i} - \bar{X})}^{2} = 9448.848$

Variance,

σ^{2} = \frac{\sum f_{i} {(x_{i} - \bar{X})}^{2}}{N} = \frac{9448.848}{2600} = 3.63

Question 3:

Find the mean, mode, S.D. and coefficient of skewness for the following data:

Year render:	10	20	30	40	50	60
No. of persons (cumulative):	15	32	51	78	97	109

Answer:

Year render	No. of persons (cumulative)	$f_{i}$	$u_{i} = \frac{x_{i} - 35}{10}$	$f_{i} u_{i}$	${u_{i}}^{2}$	$f_{i} {u_{i}}^{2}$
10	15	15	$-$ 2.5	$-$ 37.5	6.25	93.75
20	32	17	$-$ 1.5	$-$ 25.5	2.25	38.25
30	51	19	$-$ 0.5	$-$ 9.5	0.25	4.75
40	78	27	0.5	13.5	0.25	6.75
50	97	19	1.5	28.5	2.25	42.75
60	109	12	2.5	30	6.25	75
		$\sum f_{i} = N = 109$		$\sum f_{i} u_{i} = - 0.5$		$\sum f_{i} {u_{i}}^{2} = 261.25$

\bar{X} = A + h (\frac{\sum f_{i} u_{i}}{N}) = 35 + 10 (\frac{- 0.5}{109}) = 34.96 years

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 100 [\frac{261.25}{109} - \frac{0.25}{11881}] = 100 \times 2.396 = 239.6

σ = \sqrt{239.6} = 15.47 years

Coefficient of skewness = Mean

-

Mode
= 34.96

-

40
=

-

5.04

Question 4:

Find the standard deviation for the following data:

x :	3	8	13	18	23
f :	7	10	15	10	6

Answer:

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$(x_{i} - \bar{X})$	${(x_{i} - \bar{X})}^{2}$	$f_{i} {(x_{i} - \bar{X})}^{2}$
3	7	21	−9.79	95.84	670.88
8	10	80	−4.79	22.94	229.4
13	15	195	0.21	0.04	0.6
18	10	180	5.21	27.14	271.4
23	6	138	10.21	104.24	625.44
	$\sum f_{i} = 48$	$\sum f_{i} x_{i} = 614$			$\sum f_{i} {(x_{i} - \bar{X})}^{2} = 1797.32$

Variance,

σ^{2} = \frac{\sum f_{i} {(x_{i} - \bar{X})}^{2}}{\sum f_{i}} = \frac{1797.32}{48} = 37.44

S D, σ = \sqrt{37.44} = 6.12

Question 1:

Calculate the mean and S.D. for the following data:

Expenditure in Rs:	0-10	10-20	20-30	30-40	40-50
Frequency:	14	13	27	21	15

Answer:

Expenditure (Rs)	$f_{i}$	Midpoint $(x_{i})$	$f_{i} x_{i}$	$x_{i} - \bar{X}$	${(x_{i} - \bar{X})}^{2}$	$f_{i} {(x_{i} - \bar{X})}^{2}$
0−10	14	5	70	$-$ 21.1	445.21	6233.94
10−20	13	15	195	$-$ 11.1	123.21	1601.73
20−30	27	25	675	$-$ 1.1	1.21	34.67
30−40	21	35	735	8.9	79.21	1663.41
40−50	15	45	675	18.9	357.21	5358.15
	$\sum f_{i} = 90$		$\sum f_{i} x_{i} = 2350$			$\sum f_{i} {(x_{i} - \bar{X})}^{2} = 14891.9$

Mean,

\bar{X} = \frac{2350}{90} = 26.11

Variance,

σ^{2} = \frac{14891.9}{90} = 165.47

S D, σ = \sqrt{165.47} = 12.86

Question 2:

Calculate the standard deviation for the following data:

Class:	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency:	9	17	43	82	81	44	24

Answer:

Class	$f_{i}$	Midpoint $(x_{i})$	$y_{i} = \frac{x_{i} - 105}{30}$	${y_{i}}^{2}$	$f_{i} y_{i}$	$f_{i} {y_{i}}^{2}$
0−30	9	15	−3	9	−27	81
30−60	17	45	−2	4	−34	68
60−90	43	75	−1	1	−43	43
90−120	82	105	0	0	0	0
120−150	81	135	1	1	81	81
150−180	44	165	2	4	88	176
180−210	24	195	3	9	72	216
	$\sum f_{i} = N = 300$				$\sum f_{i} y_{i} = 137$	$\sum f_{i} {y_{i}}^{2} = 665$

Mean,

\bar{x} = a + h (\frac{1}{N} Σ f_{i} y_{i}) = 105 + 30 (\frac{137}{300}) = 118.7

Variance:

σ^{2} = \frac{h^{2}}{N^{2}} [N \sum f_{i} {y_{i}}^{2} - {(\sum f_{i} y_{i})}^{2}] = \frac{900}{90000} [300 \times 665 - 18769] = \frac{1}{100} [199500 - 18769] = \frac{180731}{100} = 1807.31 S D, σ = \sqrt{1807.31} = 42.51

Question 3:

Calculate the A.M. and S.D. for the following distribution:

Class:	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency:	18	16	15	12	10	5	2	1

Answer:

Class	$f_{i}$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - 35}{10}$	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
0−10	18	5	$-$ 3	$-$ 54	162
10−20	16	15	$-$ 2	$-$ 32	64
20−30	15	25	$-$ 1	$-$ 15	15
30−40	12	35	0	0	0
40−50	10	45	1	10	10
50−60	5	55	2	10	20
60−70	2	65	3	6	18
70−80	1	75	4	4	16
	$\sum f_{i} = 79$			$\sum f_{i} u_{i} = - 71$	$\sum f_{i} {u_{i}}^{2} = 305$

\bar{X} = a + h (\frac{\sum f_{i} u_{i}}{N}) = 35 + 10 (\frac{- 71}{79}) = 26.01

AM = 26.01

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 100 [\frac{305}{79} - \frac{5041}{6241}] = 305.20 σ = \sqrt{305.20} = 17.47

Question 4:

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.

Answer:

${\bar{x}}_{o} = 40, σ_{o} = 5.1, and n = 100 \Rightarrow {Σx}_{o} = 4000 Corrected sum of observations, {Σx}_{n} = 4000 - 50 + 40 = 3990 n = 100 \Rightarrow {\bar{x}}_{n} = \frac{{Σx}_{n}}{n} = 39.90 Also, SD, σ_{o} = 5.1 \Rightarrow Σ {(x_{i} - {\bar{x}}_{o})}^{2} = 2601 \Rightarrow Σ {(x_{i} - {\bar{x}}_{n})}^{2} = 2601 - {(50 - 40)}^{2} + {(40 - 39.90)}^{2} = 2601 - 100 + 0.01 = 2501.01 \Rightarrow σ_{n} = \sqrt{\frac{Σ {(x_{i} - {\bar{x}}_{n})}^{2}}{n}} = \sqrt{\frac{2501.01}{100}} = 5$

Question 5:

Calculate the mean, median and standard deviation of the following distribution:

Class-interval:	31-35	36-40	41-45	46-50	51-55	56-60	61-65	66-70
Frequency:	2	3	8	12	16	5	2	3

Answer:

Class Interval	$f_{i}$	Midpoint $x_{i}$	$u_{i} = \frac{x_{i} - 53}{4}$	u_i ²	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
31−35	2	33	$-$ 5	25	$-$ 10	50
36−40	3	38	$-$ 3.75	14.06	$-$ 11.25	42.18
41−45	8	43	$-$ 2.5	6.25	$-$ 20	50
46−50	12	48	$-$ 1.25	1.56	$-$ 15	18.72
51−55	16	53	0	0	0	0
56−60	5	58	1.25	1.56	6.25	7.8
61−65	2	63	2.5	6.25	5	12.5
66−70	3	68	3.75	14.06	11.25	42.18
	N = 51				$\sum_{i = 1}^{n} f_{i} u_{i} = - 33.75$	$\sum_{i = 1}^{n} f_{i} {u_{i}}^{2} = 223.38$

\overset{}{X} = a + h (\frac{\sum_{i = 1}^{n} f_{i} u_{i}}{N}) = 53 + 4 (\frac{- 33.75}{51}) = 50.36

σ^{2} = h^{2} (\frac{\sum_{i = 1}^{n} f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum_{i = 1}^{n} f_{i} u_{i}}{N})}^{2}) = 16 (\frac{223.38}{51} - \frac{1139.06}{2601}) = 63.07 σ = \sqrt{63.07} = 7.94

$f_{i}$	$C F$ (Cumulative frequency)
2	2
3	5
8	13
12	25
16	41
5	46
2	48
3	51

\sum f_{i} = 51 = N \frac{N}{2} = 25.5

Median class interval is 51−55.

L = 51 F = 25 f = 16 h = 4

Median = L + \frac{\frac{N}{2} - F}{f} \times h

= 51 + \frac{25.5 - 25}{16} \times 4 = 51 + \frac{0.5}{4} = 51.125

Question 6:

Find the mean and variance of frequency distribution given below:

x_i:	1 ≤ x < 3	3 ≤ x < 5	5 ≤ x < 7	7 ≤ x < 10
f_i:	6	4	5	1

Answer:

x_i	Mid-Values(y_i)	y_i²	f_i	f_i y_i	f_iy_i²
1–3	2	4	6	12	24
3–5	4	16	4	16	64
5–7	6	36	5	30	180
7–10	8.5	72.25	1	8.5	72.25
			N = $\sum_{} f_{i}$ = 16	$\sum_{} f_{i} y_{i} = 66.5$	$\sum_{} f_{i} {y_{i}}^{2} = 340.25$

Therefore,

Mean =

\frac{\sum_{} f_{i} y_{i}}{\sum_{} f_{i}} = \frac{66.5}{16} = 4.16

Variance =

(\frac{1}{N} \sum_{} f_{i} y_{i}^{2}) - {(\frac{1}{N} \sum_{} f_{i} y_{i})}^{2} = \frac{1}{16} \times 340.25 - {(\frac{1}{16} \times 66.5)}^{2} = 21.26 - 17.22 = 4.04

Question 7:

The weight of coffee in 70 jars is shown in the following table:

Weight (in grams):	200–201	201–202	202–203	203–204	204–205	205–206
Frequency:	13	27	18	10	1	1

Determine the variance and standard deviation of the above distribution. [NCERT EXEMPLAR]

Answer:

Weight (in grams)	Mid-Values $(x_{i})$	Frequency $(f_{i})$	$d_{i} = x_{i} - 202.5$	$d_{i}^{2}$	$f_{i} d_{i}$	$f_{i} d_{i}^{2}$
200–201	200.5	13	−2	4	−26	52
201–202	201.5	27	−1	1	−27	27
202–203	202.5	18	0	0	0	0
203–204	203.5	10	1	1	10	10
204–205	204.5	1	2	4	2	4
205–206	205.5	1	3	9	3	9
		N = $\sum_{} f_{i} = 70$			$\sum_{} f_{i} d_{i} = - 38$	$\sum_{} f_{i} d_{i}^{2} = 102$

Now,

Variance,

σ^{2}

= (\frac{1}{N} \sum_{} f_{i} d_{i}^{2}) - {(\frac{1}{N} \sum_{} f_{i} d_{i})}^{2} = (\frac{1}{70} \times 102) - {(\frac{1}{70} \times (- 38))}^{2} = 1.457 - 0.295 = 1.162 gm

Standard deviation, $σ$ =

\sqrt{Variance} = \sqrt{1.162} = 1.08 gm

Question 8:

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
[NCERT EXEMPLAR]

Answer:

Given:

Number of observations, n = 100

Mean, $x$ = 40

Standard deviation, $σ$ = 10

We know that

$x = \frac{\sum_{} x_{i}}{100} \Rightarrow \frac{\sum_{} x_{i}}{100} = 40 \Rightarrow \sum_{} x_{i} = 4000$

∴ Correct $\sum_{} x_{i} = 4000 - (30 + 70) + (3 + 27) = 3930$

Correct mean = $\frac{Correct \sum_{} x_{i}}{100} = \frac{3930}{100} = 39.3$

Now,

Incorrect variance, $σ^{2} = \frac{\sum_{} x_{i}^{2}}{100} - {(40)}^{2}$
$\Rightarrow \frac{\sum_{} x_{i}^{2}}{100} = 100 + 1600 = 1700 \Rightarrow \sum_{} x_{i}^{2} = 170000$

Correct $\sum_{} x_{i}^{2} = 170000 - 30^{2} - 70^{2} + 3^{2} + 27^{2} = 164939$

∴ Correct standard deviation

$= \sqrt{\frac{164939}{100} - {(39.3)}^{2}} = \sqrt{1649.39 - 1544.49} = \sqrt{104.9} = 10.24$

Question 9:

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
[NCERT EXEMPLAR]

Answer:

Given:

Number of observations, n = 10

Mean, $x$ = 45

Variance, $σ^{2}$ = 16

Now,

Incorrect mean, $x$ = 45

$\Rightarrow \frac{Incorrect \sum_{} x_{i}}{10} = 45 \Rightarrow Incorrect \sum_{} x_{i} = 450$

∴ Correct $\sum_{} x_{i} = 450 - 52 + 25 = 423$

⇒ Correct mean = $\frac{Correct \sum_{} x_{i}}{10} = \frac{423}{10} = 42.3$

Incorrect variance, $σ^{2}$ = 16

$\Rightarrow 16 = \frac{Incorrect \sum_{} x_{i}^{2}}{10} - {(45)}^{2} \Rightarrow Incorrect \sum_{} x_{i}^{2} = 10 (16 + 2025) = 20410$

∴ Correct $\sum_{} x_{i}^{2} = 20410 - {(52)}^{2} + {(25)}^{2} = 20410 - 2704 + 625 = 18331$

Now,

Correct variance = $\frac{18331}{10} - {(42.3)}^{2} = 1833.1 - 1789.29 = 43.81$

Question 10:

Calculate the mean, variance and standard deviation of the following frequency distribution.

Class:	1–10	10–20	20–30	30–40	40–50	50–60
Frequency:	11	29	18	4	5	3

Answer:

Let the assumed mean A = 25.

Class	Mid-Values $(x_{i})$	$d_{i} = x_{i} - A = x_{i} - 25$	$d_{i}^{2}$	Frequency $(f_{i})$	$f_{i} d_{i}$	$f_{i} d_{i}^{2}$
1–10	5.5	−19.5	380.25	11	−214.5	4182.75
10–20	15	−10	100	29	−290	2900
20–30	25	0	0	18	0	0
30–40	35	10	100	4	40	400
40–50	45	20	400	5	100	2000
50–60	55	30	900	3	90	2700
				N = $\sum_{} f_{i}$ = 70	$\sum_{} f_{i} d_{i}$ = −274.5	$\sum_{} f_{i} d_{i}^{2} =$ 12182.75

Mean =

A + \frac{\sum_{} f_{i} d_{i}}{\sum_{} f_{i}} = 25 + (\frac{- 274.5}{70}) = 25 - 3.92 = 21.08

Variance =

σ^{2} = (\frac{1}{N} \sum_{} f_{i} d_{i}^{2}) - {(\frac{1}{N} \sum_{} f_{i} d_{i})}^{2} = \frac{12182.75}{70} - {(\frac{- 274.5}{70})}^{2} = 174.02 - 15.37 = 158.65

Standard deviation =

σ = \sqrt{158.65} = 12.6

Question 1:

Two plants A and B of a factory show following results about the number of workers and the wages paid to them

	Plant A	Plant B
No. of workers	5000	6000
Average monthly wages	Rs 2500	Rs 2500
Variance of distribution of wages	81	100

In which plant A or B is there greater variability in individual wages?

Answer:

Variance of the distribution of wages in plant $A (σ^{2}) = 81$
Standard deviation of the distribution of wages in plant $A (σ) = 9$

Variance of the distribution of wages in plant $B (σ^{2}) = 100$
Standard deviation of the distribution of wages in plant $B (σ) = 10$
Average monthly wages in both the plants are Rs 2500.
Thus, the plant with greater value of SD will have more variability in salary.
Plant B has more variability in individual wages than plant A.

Question 2:

The means and standard deviations of heights ans weights of 50 students of a class are as follows:

	Weights	Heights
Mean	63.2 kg	63.2 inch
Standard deviation	5.6 kg	11.5 inch

Which shows more variability, heights or weights?

Answer:

$Coeffient of variations (CV) in weights = \frac{SD}{Mean} \times 100 = \frac{5.6}{63.2} \times 100 = 8.86$

$Coefficient of variations (CV) in heights = \frac{11.5}{63.2} \times 100 = 18.19$

CV in heights is greater than CV in weights.
Thus, heights will show more variability than weights.

Question 3:

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

Answer:

The coefficient of variation (CV) for the first distribution is 60.
The coefficient of variation (CV) for the second distribution is 70.
$S D (σ_{1}) = 21$
$S D (σ_{2}) = 16$
We know:
$C V = \frac{σ}{\bar{X}} \times 100$

From the above formula, we get:
$\bar{X} = \frac{σ}{C V} \times 100$

$\bar{X_{1}} = \frac{21}{60} \times 100 = 35 \bar{X_{2}} = \frac{16}{70} \times 100 = 22.86$

Question 4:

Calculate coefficient of variation from the following data:

Income (in Rs):	1000-1700	1700-2400	2400-3100	3100-3800	3800-4500	4500-5200
No. of families:	12	18	20	25	35	10

Answer:

Income (Rs)	$f_{i}$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - 3450}{700}$	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
1000−1700	12	1350	−3	−36	108
1700−2400	18	2050	−2	−36	72
2400−3100	20	2750	−1	−20	20
3100−3800	25	3450	0	0	0
3800−4500	35	4150	1	35	35
4500−5200	10	4850	2	20	40
	$\sum f_{i} = 120$			$\sum f_{i} u_{i} = - 37$	$\sum f_{i} {u_{i}}^{2} = 275$

\bar{X} = a + h (\frac{\sum f_{i} u_{i}}{N}) = 3450 + 700 (\frac{- 37}{120}) = 3234.17

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 490000 [\frac{275}{120} - \frac{1369}{14400}] = 1076332.64 σ = \sqrt{1076332.64} = 1037.46 C V = \frac{σ}{\bar{X}} \times 100 = \frac{1037.46}{3234.17} \times 100 = 32.08

Question 5:

An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

	Firm A	Firm B
No. of wage earners	586	648
Average weekly wages	Rs 52.5	Rs. 47.5
Variance of the distribution of wages	100	121

(i) Which firm A or B pays out larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?

Answer:

Average weekly wages $= \frac{Total weekly wages}{Number of workers}$
Total weekly wages = (Average weekly wages) (Numbers of workers)
Total weekly wages for firm A = Rs 52.5 $\times$ 586 = Rs 30765
Total weekly wages for firm B = Rs 47.5 $\times$ 648 = Rs 30780
(i) Firm B pays a larger amount as weekly wages.

(ii) SD (firm A) = 10
SD (firm B) = 11

$CV (firm A) = \frac{10}{52.5} \times 100 = 19.04 CV (firm B) = \frac{11}{47.5} \times 100 = 23.15$

Since CV of firm B is greater than that of firm A, firm B has greater variability in individual wages.

Question 6:

The following are some particulars of the distribution of weights of boys and girls in a class:

Number	Boys	Girls
	100	50
Mean weight	60 kg	45 kg
Variance	9	4

Which of the distributions is more variable?

Answer:

We know:
SD (boys) is 3 and SD (girls) is 2.

$CV = \frac{σ}{\bar{X}} \times 100$

$∴ CV (boys) = \frac{3}{60} \times 100 = 5$

$CV (girls) = \frac{2}{45} \times 100 = 4.44$

Since CV for boys is greater than that of girls, distribution of the weights of boys is more variable than that of girls.

Question 7:

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard	12	15	20
Deviation

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer:

We know:
$C V = \frac{σ}{\bar{X}} \times 100$

$\bar{X_{m}} = 42, σ_{m} = 12 \bar{X_{p}} = 32, σ_{p} = 15 \bar{X_{c}} = 40.9, σ_{c} = 20$
CV of mathematics marks $= \frac{12}{42} \times 100 = \frac{1200}{42} = 28.57$

CV of physics marks $= \frac{15}{32} \times 100 = \frac{1500}{32} = 46.87$

CV of chemistry marks $= \frac{20}{40.9} \times 100 = \frac{2000}{40.9} = 48.89$

Since CV of chemistry is the greatest, the variability of marks in chemistry is the highest and that of mathematics is the lowest.

Question 8:

From the data given below state which group is more variable, G₁ or G₂?

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Group G₁	9	17	32	33	40	10	9
Group G₂	10	20	30	25	43	15	7

Answer:

Marks	$f_{i}$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - 45}{10}$	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
10−20	9	15	−3	−27	81
20−30	17	25	−2	−34	68
30−40	32	35	−1	−32	32
40−50	33	45	0	0	0
50−60	40	55	1	40	40
60−70	10	65	2	20	40
70−80	9	75	3	27	81
	N=150			$\sum f_{i} u_{i} = - 6$	$\sum f_{i} {u_{i}}^{2} = 342$

h = 5, a = 45

For group 1:

\bar{X} = a + h (\frac{\sum f_{i} u_{i}}{N}) = 45 + 10 (\frac{- 6}{150}) = 44.6

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 100 [\frac{342}{150} - \frac{36}{22500}] = 227.84 σ = \sqrt{227.84} = 15.09

Marks	$f_{i}$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - 45}{10}$	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
10−20	10	15	−3	−30	90
20−30	20	25	−2	−40	80
30−40	30	35	−1	−30	30
40−50	25	45	0	0	0
50−60	43	55	1	43	43
60−70	15	65	2	30	60
70−80	7	75	3	21	63
	$\sum f_{i} = 150$			$\sum f_{i} u_{i} = - 6$	$\sum f_{i} {u_{i}}^{2} = 366$

For group 2:

\bar{X} = a + h (\frac{\sum f_{i} u_{i}}{N}) = 45 + 10 (\frac{- 6}{150}) = 44.6

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 100 [\frac{366}{150} - \frac{36}{22500}] = 243.84 σ = \sqrt{243.84} = 15.62

Mean of both the groups are same and SD of group 2 is greater than that of group 1.
So, group 2 will be more variable.

Question 9:

Find the coefficient of variation for the following data:

Size (in cms):	10-15	15-20	20-25	25-30	30-35	35-40
No. of items:	2	8	20	35	20	15

Answer:

Size (cm)	$f_{i}$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - 27.5}{5}$	$f_{i} u_{i}$	$f_{i} {u_{i}}^{2}$
10−15	2	12.5	$-$ 3	$-$ 6	18
15−20	8	17.5	$-$ 2	$-$ 16	32
20−25	20	22.5	$-$ 1	$-$ 20	20
25−30	35	27.5	0	0	0
30−35	20	32.5	1	20	20
35−40	15	37.5	2	30	60
	$\sum f_{i} = N = 100$			$\sum f_{i} u_{i} = 8$	$\sum f_{i} {u_{i}}^{2} = 150$

Here,

h = 5, a = 27.5

\bar{X} = a + h (\frac{\sum f_{i} u_{i}}{N}) = 27.5 + 5 (\frac{8}{100}) = 27.9

σ^{2} = h^{2} [\frac{\sum f_{i} {u_{i}}^{2}}{N} - {(\frac{\sum f_{i} u_{i}}{N})}^{2}] = 25 [\frac{150}{100} - \frac{64}{10000}] = 37.34

σ = \sqrt{37.34} = 6.11

We know that

C V = \frac{σ}{\bar{X}} \times 100 = \frac{6.11}{27.9} \times 100 = 21.9

Question 10:

From the prices of shares X and Y given below: find out which is more stable in value:

X:	35	54	52	53	56	58	52	50	51	49
Y:	108	107	105	105	106	107	104	103	104	101

Answer:

Let A_x = 51

$x_{i}$	$d_{i} = x_{i} - 51$	${d_{i}}^{2}$
35	$-$ 16	256
54	3	9
52	1	1
53	2	4
56	5	25
58	7	49
52	1	1
50	$-$ 1	1
51	0	0
49	$-$ 2	4
	$\sum d_{i} = 0$	$\sum {d_{i}}^{2} = 350$

Here, we have

n = 10, \bar{X} = 51 σ^{2} = \frac{\sum {d_{i}}^{2}}{n} - {(\frac{\sum d_{i}}{n})}^{2} = \frac{350}{10} - {(\frac{0}{10})}^{2} = 35 - 0 = 35

σ = \sqrt{35} = 5.91

{CV}_{x} = \frac{5.91}{51} \times 100 = 11.58

Let A_y =105

$x_{i}$	$d_{i} = x_{i} - 105$	${d_{i}}^{2}$
108	3	9
107	2	4
105	0	0
105	0	0
106	1	1
107	2	4
104	$-$ 1	1
103	$-$ 2	4
104	$-$ 1	1
101	$-$ 4	16
	$\sum d_{i} = 0$	$\sum {d_{i}}^{2} = 40$

n = 10, \bar{Y} = 105 σ^{2} = \frac{\sum {d_{i}}^{2}}{n} - {(\frac{\sum d_{i}}{n})}^{2} = \frac{40}{10} - {(\frac{0}{10})}^{2} = 4 - 0 = 4

σ = \sqrt{4} = 2

{CV}_{y} = \frac{2}{105} \times 100 = 1.90

Since CV of prices of share Y is lesser than that of X, prices of shares Y are more stable.

Question 11:

Life of bulbs produced by two factories A and B are given below:

Length of life (in hours):	550–650	650–750	750–850	850–950	950–1050
Factory A: (Number of bulbs)	10	22	52	20	16
Factory B: (Number of bulbs)	8	60	24	16	12

The bulbs of which factory are more consistent from the point of view of length of life? [NCERT EXEMPLAR]

Answer:

For factory A

Let the assumed mean A = 800 and h = 100.

Length of Life (in hours)	Mid-Values $(x_{i})$	$u_{i} = \frac{x_{i} - 800}{100}$	$u_{i}^{2}$	Number of bulbs $(f_{i})$	$f_{i} u_{i}$	$f_{i} u_{i}^{2}$
550–650	600	−2	4	10	−20	40
650–750	700	−1	1	22	−22	22
750–850	800	0	0	52	0	0
850–950	900	1	1	20	20	20
950–1050	1000	2	4	16	32	64
				$\sum_{} f_{i} = 120$	$\sum_{} f_{i} u_{i} = 10$	$\sum_{} f_{i} u_{i}^{2} = 146$

Mean,

X_{A} = A + h (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}}) = 800 + 100 \times \frac{10}{120} = 808.33

Standard deviation,

σ_{A} = h \sqrt{(\frac{\sum_{} f_{i} u_{i}^{2}}{\sum_{} f_{i}}) - {(\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})}^{2}} = 100 \sqrt{(\frac{146}{120}) - {(\frac{10}{120})}^{2}} = 100 \times 1.0998 = 109.98

∴ Coefficient of variation =

\frac{σ_{A}}{X_{A}} \times 100 = \frac{109.98}{808.33} \times 100 = 13.61

For factory B

Let the assumed mean A = 800 and h = 100.

Length of Life (in hours)	Mid-Values $(x_{i})$	$u_{i} = \frac{x_{i} - 800}{100}$	$u_{i}^{2}$	Number of bulbs $(f_{i})$	$f_{i} u_{i}$	$f_{i} u_{i}^{2}$
550–650	600	−2	4	8	−16	32
650–750	700	−1	1	60	−60	60
750–850	800	0	0	24	0	0
850–950	900	1	1	16	16	16
950–1050	1000	2	4	12	24	48
				$\sum_{} f_{i} = 120$	$\sum_{} f_{i} u_{i} = - 36$	$\sum_{} f_{i} u_{i}^{2} = 156$

Mean,

X_{B} = A + h (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}}) = 800 + 100 \times \frac{(- 36)}{120} = 800 - 30 = 770

Standard deviation,

σ_{B} = h \sqrt{(\frac{\sum_{} f_{i} u_{i}^{2}}{\sum_{} f_{i}}) - {(\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})}^{2}} = 100 \sqrt{(\frac{156}{120}) - {(\frac{- 36}{120})}^{2}} = 100 \times 1.1 = 110

∴ Coefficient of variation =

\frac{σ_{B}}{X_{B}} \times 100 = \frac{110}{770} \times 100 = 14.29

Since the coefficient of variation of factory B is greater than the coefficient of variation of factory A, therefore, factory B has more variability than factory A. This means bulbs of factory A are more consistent from the point of view of length of life.

Question 12:

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:

Ravi:	25	50	45	30	70	42	36	48	35	60
Hashina:	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent? [NCERT EXEMPLAR]

Answer:

For Ravi

Marks $(x_{i})$	$d_{i} = x_{i} - 45$	$d_{i}^{2}$
25	−20	400
50	5	25
45	0	0
30	−15	225
70	25	625
42	−3	9
36	−9	81
48	3	9
35	−10	100
60	15	225
	$\sum_{} d_{i} = - 9$	$\sum_{} d_{i}^{2} = 1699$

Mean,

X_{R} = A + \frac{\sum_{} d_{i}}{10} = 45 + \frac{(- 9)}{10} = 44.1

Standard deviation,

σ_{R} = \sqrt{\frac{\sum_{} d_{i}^{2}}{10} - {(\frac{\sum_{} d_{i}}{10})}^{2}} = \sqrt{\frac{1699}{10} - {(\frac{- 9}{10})}^{2}} = \sqrt{169.09} = 13.003

Coefficicent of variation =

\frac{σ_{R}}{X_{R}} \times 100 = \frac{13.003}{44.1} \times 100 = 29.49

For Hashina

Marks $(x_{i})$	$d_{i} = x_{i} - 55$	$d_{i}^{2}$
10	−45	2025
70	15	625
50	−5	25
20	−35	1225
95	40	1600
55	0	0
42	−13	169
60	5	25
48	−7	49
80	25	625
	$\sum_{} d_{i} = - 20$	$\sum_{} d_{i}^{2} = 6368$

Mean,

X_{H} = A + \frac{\sum_{} d_{i}}{10} = 55 + \frac{(- 20)}{10} = 53

Standard deviation,

σ_{H} = \sqrt{\frac{\sum_{} d_{i}^{2}}{10} - {(\frac{\sum_{} d_{i}}{10})}^{2}} = \sqrt{\frac{6368}{10} - {(\frac{- 20}{10})}^{2}} = \sqrt{632.8} = 25.16

Coefficicent of variation =

\frac{σ_{H}}{X_{H}} \times 100 = \frac{25.16}{53} \times 100 = 47.47

Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.

Question 1:

For a frequency distribution mean deviation from mean is computed by
(a) M.D. = $\frac{Σ f}{Σ f |d|}$

(b) M.D. = $\frac{Σ d}{Σ f}$

(c) M.D. = $\frac{Σ f d}{Σ f}$

(d) M.D. = $\frac{Σ f |d|}{Σ f}$

Answer:

$(d) MD = \frac{\sum_{}^{} f |d|}{\sum_{}^{} f}$

Question 2:

For a frequency distribution standard deviation is computed by applying the formula
(a) $σ = \sqrt{\frac{Σ f d^{2}}{Σ f} - {(\frac{Σ f d}{Σ f})}^{2}}$

(b) $σ = \sqrt{{(\frac{Σ f d}{Σ f})}^{2} - \frac{Σ f d^{2}}{Σ f}}$

(c) $σ = \sqrt{\frac{Σ f d^{2}}{Σ f} - \frac{Σ f d}{Σ f}}$

(d) $\sqrt{{(\frac{Σ f d}{Σ f})}^{2} - \frac{Σ f d^{2}}{Σ f}}$

Answer:

$(a) σ = \sqrt{\frac{\sum_{}^{} f d^{2}}{\sum_{}^{} f} - {(\frac{\sum_{}^{} f d}{\sum_{}^{} f})}^{2}}$

Question 3:

If v is the variance and σ is the standard deviation, then
(a) $v = \frac{1}{σ^{2}}$

(b) $v = \frac{1}{σ}$

(c) v = σ²

(d) v² = σ

Answer:

$(c) v = σ^{2}$

The variance is the square of the standard deviation.

Question 4:

The mean deviation from the median is
(a) equal to that measured from another value
(b) maximum if all observations are positive
(c) greater than that measured from any other value.
(d) less than that measured from any other value.

Answer:

(d) less than that measured from any other value.

In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.

Question 5:

If n = 10, $X = 12$ and $Σ x_{i}^{2} = 1530$ , then the coefficient of variation is
(a) 36%
(b) 41%
(c) 25%
(d) none of these

Answer:

(c) 25%

Standard deviation is expressed in the following manner:
$σ = \sqrt{\frac{1}{n} \sum_{i} {x_{i}}^{2} - {(X)}^{2}} = \sqrt{\frac{1530}{10} - {(12)}^{2}} = \sqrt{9} = 3$

$C V = \frac{σ}{X} \times 100 = \frac{3}{12} \times 100 = 25 %$

Question 6:

The standard deviation of the data:

x:	1	a	a²	....	aⁿ
f:	ⁿC₀	ⁿC₁	ⁿC₂	....	ⁿC_n

is
(a)

{(\frac{1 + a^{2}}{2})}^{n} - {(\frac{1 + a}{2})}^{n}

(b)

{(\frac{1 + a^{2}}{2})}^{2 n} - {(\frac{1 + a}{2})}^{n}

(c)

{(\frac{1 + a}{2})}^{2 n} - {(\frac{1 + a^{2}}{2})}^{n}

(d) none of these

Answer:

(d) None of these

x_i	f_i	f_ix_i	${x_{i}}^{2}$	$f_{i} {x_{i}}^{2}$
1	${}^{n}C_{0}$	${}^{n}C_{0}$	1	1
a	${}^{n}C_{1}$	a ${}^{n}C_{1}$	a²	a² ${}^{n}C_{1}$
a²	${}^{n}C_{2}$	a² ${}^{n}C_{2}$	a⁴	a⁴ ${}^{n}C_{2}$
a³	${}^{n}C_{3}$	a³ ${}^{n}C_{3}$	a⁶	a⁶ ${}^{n}C_{3}$
: : : :	: : :	: : :	: : :	: : : :
aⁿ	${}^{n}C_{n}$	aⁿ ${}^{n}C_{n}$	a²ⁿ	a²ⁿ ${}^{n}C_{n}$
	$\sum_{i = 1}^{n} f_{i} = 2^{n}$	$\sum_{i = 1}^{n} f_{i} x_{i} = {(1 + a)}^{n}$		$\sum_{i = 1}^{n} f_{i} {x_{i}}^{2} = (1 + a^{2})^{n}$

Number of terms, N = \underset{}{\sum_{i = 1}^{n}} f_{i} = 2^{n} \underset{}{\sum_{i = 1}^{n} f_{i}} x_{i} = {}^{n}C_{0} + a {}^{n}C_{1} + a^{2} {}^{n}C_{2} + . . . + a^{n} {}^{_{n}}C_{n} = {(1 + a)}^{n} X = \frac{\underset{}{\sum_{i = 1}^{n} f_{i}} x_{i}}{N} = \frac{{(1 + a)}^{n}}{2^{n}} \underset{}{\sum_{i = 1}^{n} f_{i}} {x_{i}}^{2} = {(1 + a^{2})}^{n} σ^{2} = Variance (X) = \underset{}{\frac{1}{N} \sum_{i = 1}^{n} f_{i}} {x_{i}}^{2} - {(\frac{\underset{}{\sum_{i = 1}^{n} f_{i}} x_{i}}{N})}^{2} = \frac{{(1 + a^{2})}^{n}}{2^{n}} - {[\frac{{(1 + a)}^{n}}{2^{n}}]}^{2} = {[\frac{1 + a^{2}}{2}]}^{n} - {[\frac{1 + a}{2}]}^{2 n} σ = \sqrt{Variance (X)} = \sqrt[]{{[\frac{1 + a^{2}}{2}]}^{n} - {[\frac{1 + a}{2}]}^{2 n}}

Question 7:

The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is
(a) $\frac{(n + 1) d}{2 n + 1}$

(b) $\frac{n d}{2 n + 1}$

(c) $\frac{n (n + 1) d}{2 n + 1}$

(d) $\frac{(2 n + 1) d}{n (n + 1)}$

Answer:

$(c) \frac{n (n + 1) d}{2 n + 1}$

$x_{i}$	$\|x_{i} - X\| = \|x_{i} - (a + n d)\|$
a	nd
a + d	(n-1)d
a + 2d	(n-2)d
a + 3d	(n-3)d
:	:
:	:
a + (n - 1)d	d
a + nd	0
a + (n+1)d	d
:	:
:	:
a + 2nd	nd
$\sum_{} x_{i} = (2 n + 1) (a + n d)$	$\sum_{} \|x_{i} - X\| = n (n + 1) d$

There are 2 n + 1 terms . \Rightarrow N = 2 n + 1 \sum_{} x_{i} = a + a + d + a + 2 d + a + 3 d + . . . + a + 2 n d = (2 n + 1) a + d (1 + 2 + 3 + . . . + 2 n) [a + a + a + . . . (2 n + 1) times = (2 n + 1) a] = (2 n + 1) a + \frac{2 n (2 n + 1) d}{2} [Sum of the first n natural numbers is \frac{n (n + 1)}{2}, but here we are considering the first 2 n numbers .] = (2 n + 1) a + (2 n + 1) n d = (2 n + 1) (a + n d) X = \frac{(2 n + 1) (a + n d)}{(2 n + 1)} = a + n d \sum_{} |x_{i} - X| = n d + (n - 1) d + (n - 2) d + . . . + d + 0 + d + 2 d + 3 d + . . . + n d = d (n + (n - 1) + (n - 2) + . . . + 1) + 0 + d (1 + 2 + 3 + . . . + n) = \frac{d n (n + 1)}{2} + \frac{d n (n + 1)}{2} \{∵ 1 + 2 + 3 + . . . + n = \frac{n (n + 1)}{2}\} = n (n + 1) d Mean deviation about the mean = \frac{\sum_{} |x_{i} - X|}{N} = \frac{n (n + 1) d}{(2 n + 1)}

Question 8:

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
(a) 8.6
(b) 6.4
(c) 10.6
(d) 7.6

Answer:

Disclaimer: No option is matching the answer.

$N = 10 X = \frac{38 + 70 + 48 + 34 + 42 + 55 + 63 + 46 + 54 + 44}{10} = \frac{494}{10} = 49.4$

x_i	d_i = $\|x_{i} - 49.4\|$
34	15.4
38	11.4
42	7.4
44	5.4
46	3.4
48	1.4
54	4.6
55	5.6
63	13.6
70	20.6
	$\sum_{i =}^{n} d_{i} = 88.8$

Mean deviation from the mean = \frac{88.8}{10} = 8.88

Question 9:

The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
(a) 25
(b) 5
(c) 1.2
(d) 0

Answer:

(c) 1.2

$Mean (X) = \frac{3 + 4 + 5 + 6 + 7}{5} = \frac{25}{5} = 5$
Taking the absolute value of deviation of each term from the mean, we get:

$MD = \frac{|(3 - 5)| + |(4 - 5)| + |(5 - 5)| + |(6 - 5)| + |(7 - 5)|}{5} = \frac{2 + 1 + 0 + 1 + 2}{5} = \frac{6}{5} = 1.2$

Question 10:

The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
(a) 10 %
(b) 40 %
(c) 50 %
(d) none of these

Answer:

(a) 10%

$We have : X = 50, n = 10 \sum_{i = 1}^{10} {(x_{i} - X)}^{2} = 250 ∴ SD = \sqrt{Variance of X} = \sqrt{\frac{\sum_{i = 1}^{10} {(x_{i} - X)}^{2}}{n}} = \sqrt{\frac{250}{10}} = 5$

Using $CV = \frac{σ}{X} \times 100$

$\Rightarrow CV = \frac{5}{50} \times 100 = 10 %$

Question 11:

Let x₁, x₂, ..., x_n be values taken by a variable X and y₁, y₂, ..., y_n be the values taken by a variable Y such that y_i = ax_i + b, i = 1, 2,..., n. Then,
(a) Var (Y) = a² Var (X)
(b) Var (X) = a² Var (Y)
(c) Var (X) = Var (X) + b
(d) none of these

Answer:

$(a) V a r (Y) = a^{2} V a r (X)$

$Var (X) = \frac{\sum_{i = 1}^{n} (x_{i} - \overset{}{\bar{X}})^{2}}{n} where Mean (X) = \frac{\sum_{i = 1}^{n} x_{i}}{n} Var (Y) = \frac{\sum_{i = 1}^{n} (y_{i} - Y)^{2}}{n} and Y = \frac{\sum_{i = 1}^{n} y_{i}}{n} We have, y_{i} = a x_{i} + b Y = \frac{\sum_{i = 1}^{n} y_{i}}{n} = \frac{\sum_{i = 1}^{n} a x_{i} + b}{n} = \frac{a \sum_{i = 1}^{n} x_{i}}{n} + \frac{n b}{n} = a X + b V a r (Y) = \frac{\sum_{i = 1}^{n} {(y_{i} - Y)}^{2}}{n} = \frac{\sum_{i = 1}^{n} {\{a x_{i} + b - (a X + b)\}}^{2}}{n} = \frac{\sum_{i = 1}^{n} (a x_{i} - a X)^{2}}{n} = a^{2} \frac{\sum_{i = 1}^{n} (x_{i} - X)^{2}}{n} = a^{2} V a r (X)$

Question 12:

If the standard deviation of a variable X is σ, then the standard deviation of variable $\frac{a X + b}{c}$ is
(a) a σ

(b) $\frac{a}{c} σ$

(c) $|\frac{a}{c}| σ$

(d) $\frac{a σ + b}{c}$

Answer:

(c) $|\frac{a}{c}| σ$

$Y = \frac{a X + b}{c} Y = \frac{\sum y_{i}}{n} = \frac{\frac{a \sum X + n b}{c}}{n} = \frac{a \sum X}{n c} + \frac{n b}{n c} = \frac{a \bar{X}}{c} + \frac{b}{c}$

$Var (X) = \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = σ^{2} Var (Y) = \frac{\sum {(y_{i} - \bar{Y})}^{2}}{n} = \frac{\sum {(\frac{a X}{c} + \frac{b}{c} - \frac{a}{c} \bar{X} - \frac{b}{c})}^{2}}{n} = \frac{\sum {(\frac{a X}{c} - \frac{a}{c} \bar{X})}^{2}}{n} = {(\frac{a}{c})}^{2} \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = {(\frac{a}{c})}^{2} σ^{2} SD (σ) = \sqrt{{(\frac{a}{c})}^{2} σ^{2}} = |\frac{a}{c}| σ$

Question 13:

If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
(a) −4
(b) −8
(c) 8
(d) 4

Answer:

(d) 4

If a set of observations, with SD $σ$ , are multiplied with a non-zero real number a, then SD of the new observations will be $|a| σ .$
Dividing the set of observations by − 2 is same as multiplying the observations by $\frac{1}{- 2}$ .

$New S . D . = |- \frac{1}{2}| \times 8 = \frac{8}{2} = 4$

Question 14:

If two variates X and Y are connected by the relation $Y = \frac{a X + b}{c}$ , where a, b, c are constants such that ac < 0, then
(a) $σ_{Y} = \frac{a}{c} σ_{X}$

(b) $σ_{Y} = - \frac{a}{c} σ_{X}$

(c) $σ_{Y} = \frac{a}{c} σ_{X} + b$

(d) none of these

Answer:

$(b) σ_{Y} = - \frac{a}{c} σ_{X}$

$Y = \frac{a X + b}{c} Y = \frac{\sum_{i = 1}^{n} \frac{a X + b}{c}}{n} = \frac{\frac{a \sum_{i = 1}^{n} X + n b}{c}}{n} = \frac{\frac{a}{c} \sum_{i = 1}^{n} X}{n} + \frac{b}{c} = \frac{a X}{c} + \frac{b}{c} We know : V a r (X) = \frac{\sum_{i = 1}^{n} {(x_{i} - X)}^{2}}{n} = σ^{2} V a r (Y) = \frac{\sum_{i = 1}^{n} (y_{i} - Y)^{2}}{n} = \frac{\sum_{i = 1}^{n} {(\frac{a X}{c} + \frac{b}{c} - \frac{a}{c} X - \frac{b}{c})}^{2}}{n} = \frac{\sum_{i = 1}^{n} {(\frac{a X}{c} - \frac{a}{c} X)}^{2}}{n} = {(\frac{a}{c})}^{2} \frac{\sum_{i = 1}^{n} {(x_{i} - X)}^{2}}{n} = {(\frac{a}{c})}^{2} σ^{2} SD of Y (σ_{y}) = \sqrt{{(\frac{a}{c})}^{2} σ^{2}} = |\frac{a}{c}| σ$

$a c < 0 \Rightarrow a < 0 or c < 0 ∴ |\frac{a}{c}| = - \frac{a}{c}$

$\Rightarrow σ_{Y} = - \frac{a}{c} σ_{X}$

Question 15:

If for a sample of size 60, we have the following information $\sum_{} x_{i}^{2} = 18000$ and $\sum_{} x_{i} = 960$ , then the variance is

(a) 6.63 (b) 16 (c) 22 (d) 44

Answer:

Given: $\sum_{} x_{i}^{2} = 18000$ , $\sum_{} x_{i} = 960$ and n = 60

∴ Variance

$= \frac{\sum_{} x_{i}^{2}}{n} - {(\frac{\sum_{} x_{i}}{n})}^{2} = \frac{18000}{60} - {(\frac{960}{60})}^{2} = 300 - 256 = 44$

Hence, the correct answer is option (d).

Question 16:

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

(a) s (b) ks (c) s + k (d) $\frac{s}{k}$

Answer:

The given observations are a, b, c, d, e.

Mean = m = $\frac{a + b + c + d + e}{5}$

$\Rightarrow \sum_{} x_{i} = a + b + c + d + e = 5 m$    .....(1)

Standard deviation, s = $\sqrt{\frac{\sum_{} x_{i}^{2}}{5} - m^{2}}$

Now, consider the observations a + k, b + k, c + k, d + k, e + k.

New mean $= \frac{(a + k) + (b + k) + (c + k) + (d + k) + (e + k)}{5}$

                $= \frac{a + b + c + d + e + 5 k}{5} = \frac{5 m + 5 k}{5} = m + k$

∴ New standard deviation

$= \sqrt{\frac{\sum_{} {(x_{i} + k)}^{2}}{5} - {(m + k)}^{2}} = \sqrt{\frac{\sum_{} (x_{i}^{2} + k^{2} + 2 x_{i} k)}{5} - (m^{2} + k^{2} + 2 m k)} = \sqrt{\frac{\sum_{} x_{i}^{2}}{5} + \frac{\sum_{} k^{2}}{5} + \frac{\sum_{} 2 x_{i} k}{5} - (m^{2} + k^{2} + 2 m k)} = \sqrt{\frac{\sum_{} x_{i}^{2}}{5} - m^{2} + \frac{5 k^{2}}{5} - k^{2} + \frac{2 k \sum_{} x_{i}}{5} - 2 m k}$
$= \sqrt{\frac{\sum_{} x_{i}^{2}}{5} - m^{2} + \frac{2 k \times 5 m}{5} - 2 m k} [Using (1)] = \sqrt{\frac{\sum_{} x_{i}^{2}}{5} - m^{2}} = s$

Hence, the correct answer is option (a).

Question 17:

The standard deviation of first 10 natural numbers is

(a) 5.5 (b) 3.87 (c) 2.97 (d) 2.87

Answer:

We know that the standard deviation of first n natural number is $\sqrt{\frac{n^{2} - 1}{12}}$ .

∴ Standard deviation of first 10 natural numbers

$= \sqrt{\frac{10^{2} - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25} = 2.87$

Hence, the correct answer is option (d).

Question 18:

Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is

(a) 8.25 (b) 6.5 (c) 3.87 (d) 2.87

Answer:

The first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Multiplying each number by −1, we get

−1, −2, −3, −4, −5, −6, −7, −8, −9, −10

Adding 1 to each of these numbers, we get

0, −1, −2, −3, −4, −5, −6, −7, −8, −9

Now,

$\sum_{} x_{i} = 0 + (- 1) + (- 2) + (- 3) + (- 4) + (- 5) + (- 6) + (- 7) + (- 8) + (- 9) = - 45$

$\sum_{} x_{i}^{2} = 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285$

∴ Variance of the obtained numbers

$= \frac{\sum_{} x_{i}^{2}}{10} - {(\frac{\sum_{} x_{i}}{10})}^{2} = \frac{285}{10} - {(\frac{- 45}{10})}^{2} = 28.5 - 20.25 = 8.25$

Hence, the correct answer is option (a).

Question 19:

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

(a) 6.5 (b) 2.87 (c) 3.87 (d) 8.25

Answer:

The given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

If 1 is added to each number, then the new numbers obtained are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Now,

$\sum_{} x_{i} = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65$

$\sum_{} x_{i}^{2} = 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505$

∴ Variance of the numbers so obtained

$= \frac{\sum_{} x_{i}^{2}}{10} - {(\frac{\sum_{} x_{i}}{10})}^{2} = \frac{505}{10} - {(\frac{65}{10})}^{2} = 50.5 - 42.25 = 8.25$

Hence, the correct answer is option (d).

Question 20:

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is

(a) 50,000 (b) 250,000 (c) 252500 (d) 255000

Answer:

Let $x$ and $σ$ be the mean and standard deviation of 100 observations, respectively.

$∴ x = 50, σ = 5$ and n = 100

Mean, $x = 50$

$\Rightarrow \frac{\sum_{} x_{i}}{100} = 50 \Rightarrow \sum_{} x_{i} = 5000 . . . . . (1)$

Now,

Standard deviation, $σ = 5$

$\Rightarrow \sqrt{\frac{\sum_{} x_{i}^{2}}{100} - {(\frac{\sum_{} x_{i}}{100})}^{2}} = 5 \Rightarrow \frac{\sum_{} x_{i}^{2}}{100} - {(\frac{5000}{100})}^{2} = 25 [From (1)] \Rightarrow \frac{\sum_{} x_{i}^{2}}{100} = 25 + 2500 = 2525 \Rightarrow \sum_{} x_{i}^{2} = 252500$

Thus, the sum of all squares of all the observations is 252500.

Hence, the correct answer is option (c).

Question 21:

Let x₁, x₂, ..., x_n be n observations. Let $y_{i} = a x_{i} + b$ for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of $x_{i}' s$ is 48 and their standard deviation is 12, the mean of $y_{i}' s$ is 55 and standard deviation of $y_{i}' s$ is 15, the values of a and b are

(a) a = 1.25, b = −5 (b) a = −1.25, b = 5 (c) a = 2.5, b = −5 (d) a = 2.5, b = 5

Answer:

It is given that $y_{i} = a x_{i} + b$ for i = 1, 2, 3, ..., n, where a and b are constants.

$x_{i}$ = 48 and $σ_{x_{i}} = 12$

$y_{i} = 55$ and $σ_{y_{i}} = 15$

$y_{i} = a x_{i} + b \Rightarrow \frac{\sum_{} y_{i}}{n} = \frac{\sum_{} (a x_{i} + b)}{n} \Rightarrow \frac{\sum_{} y_{i}}{n} = a \frac{\sum_{} x_{i}}{n} + \frac{\sum_{} b}{n} \Rightarrow y_{i} = a x_{i} + b \Rightarrow 55 = 48 a + b . . . . . (1)$

Now,

Standard deviation of y_i= Standard deviation of $a x_{i} + b$

$\Rightarrow σ_{y_{i}} = a \times σ_{x_{i}} \Rightarrow 15 = 12 a \Rightarrow a = \frac{15}{12} = 1.25$

Putting a = 1.25 in (1), we get

$b = 55 - 48 \times 1.25 = 55 - 60 = - 5$

Thus, the values of a and b are 1.25 and −5, respectively.

Hence, the correct answer is option (a).

Question 22:

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is

(a) 2 (b) 2.57 (c) 3 (d) 3.57

Answer:

The given observations are 3, 10, 10, 4, 7, 10, 5.

∴ Mean, $x = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7$

Now,

Mean deviation from mean, MD

$= \frac{\sum_{} |x_{i} - 7|}{7}$

$= \frac{|3 - 7| + |10 - 7| + |10 - 7| + |4 - 7| + |7 - 7| + |10 - 7| + |5 - 7|}{7} = \frac{4 + 3 + 3 + 3 + 0 + 3 + 2}{7} = \frac{18}{7} = 2.57$

Hence, the correct answer is (b).

Question 23:

The mean deviation for n observations $x_{1}, x_{2}, . . ., x_{n}$ from their mean $X$ is given by

(a) $\sum_{i = 1}^{n} (x_{i} - X)$ (b) $\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - X)$ (c) $\sum_{i = 1}^{n} {(x_{i} - X)}^{2}$ (d) $\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}$

Answer:

The mean deviation for n observations $x_{1}, x_{2}, . . ., x_{n}$ from their mean $X$ is $\frac{1}{n} \sum_{i = 1}^{n} |x_{i} - X|$ .

Disclaimer: There is some printing error in option (b) given in the question. The answer would be option (b) if it given as $\frac{1}{n} \sum_{i = 1}^{n} |x_{i} - X|$ .

Question 24:

Let $x_{1}, x_{2}, . . ., x_{n}$ be n observations and $X$ be their arithmetic mean. The standard deviation is given by

(a) $\sum_{i = 1}^{n} {(x_{i} - X)}^{2}$ (b) $\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}$ (c) $\sqrt{\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}}$ (d) $\sqrt{\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - X^{2}}$

Answer:

It is given that $x_{1}, x_{2}, . . ., x_{n}$ are n observations and $X$ is their arithmetic mean.

The standard deviation of given observations is $\sqrt{\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}}$ .

Also,

$\sqrt{\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - X)}^{2}}$ = $\sqrt{\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - X^{2}}$

Hence, the correct answers are options (c) and (d).

Disclaimer: For option (c) to be the only correct answer, option (d) should be different from the given value.

Question 25:

The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is

(a) 6 (b) $\sqrt{6}$ (c) $\frac{52}{7}$ (d) $\sqrt{\frac{52}{7}}$

Answer:

The given observations are 6, 5, 9, 13, 12, 8, 10.

Now,

$\sum_{} x_{i} = 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63$

$\sum_{} x_{i}^{2} = 36 + 25 + 81 + 169 + 144 + 64 + 100 = 619$

∴ Standard deviation of the observations, $σ$

$= \sqrt{\frac{1}{N} \sum_{} x_{i}^{2} - {(\frac{1}{N} \sum_{} x_{i})}^{2}} = \sqrt{\frac{1}{7} \times 619 - {(\frac{1}{7} \times 63)}^{2}} = \sqrt{\frac{619}{7} - 81} = \sqrt{\frac{619 - 567}{7}} = \sqrt{\frac{52}{7}}$

Hence, the correct answer is option (d).

Question 26:

The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is
(a) 2.23
(b) 2.57
(c) 3.23
(d) 3.57

Answer:

Given data is 2, 9, 9, 3, 6, 9, 4
By arranging the observations in ascending order of magnitude,
We obtain 2, 3, 4, 6, 9, 9, 9
Mean of given observations

$\begin{array}{rcl} \bar{X} & = & \frac{2 + 3 + 4 + 6 + 9 + 9 + 9}{7} \\ = & \frac{42}{7} \\ \bar{X} & = & 6 \end{array}$

*x_i*	$\|d_{i}\| = \|x_{i} - \bar{x}\| = \|x_{i} - 6\|$
2 3 4 6 9 9 9 Total	4 3 2 0 3 3 3 $\sum_{} d_{i} = 18$

Then Mean deviation is \frac{1}{x} \sum_{} |d_{i}| = \frac{1}{7} \times 18

i.e Mean deviation = 2.57

Hence, the correct answer is option B.

Question 27:

Variance of the data 2, 4, 5, 6, 8,17 is 23.33. The variance of 4, 8, 10, 12, 16, 34 will be
(a) 23.33
(b) 25.33
(c) 46.66
(d) 93.32

Answer:

Given ; Variance of 2, 4, 5, 6, 8, 17 is 23.33 = var(X)(say)
Using result ;
If variance of x₁, x₂,..........., x_n is var(X)
Then variance of ax₁, ax₂,............., ax_n is a² var(X)
Then variance of 4, 8, 10, 12, 16, 32
i.e 2(2), 2(4), 2(5), 2(6), 2(8), 2(17)
is (2)² var(X)
= 2² × 23.33
= 4 × 23.33 = 93.32

Hence, the correct answer is option D.

Question 28:

A set of n values x₁, x₂, ...., x_n has standard deviation σ. The standard deviation of n values x₁+ k, x₂+ k,..., x_n+ k will be
(a) σ
(b) σ + k
(c) σ – k
(d) k^σ

Answer:

Given set of n observations x₁, x₂,....., x_n has standard deviation σ.
i.e x₁, x₂,........., x_n has variance = σ²
i.e x₁ + k, x₂ + k₁,......... x_n + k has same variance σ²
Hence standard deviation of x₁ + k, x₂ + k₁,,........, x_n + k is σ.

Hence, the correct answer is option A.

Question 29:

Let x₁, x₂, x₃, x₄, x₅ be the observations with mean m and standard deviations. The standard deviation of the observation kx₁, kx₂, kx₃, kx₄, kx₅, is
(a) k + s

(b) $\frac{s}{k}$

c) $|k| s$

(d) s

Answer:

Given, x₁, x₂, x₃, x₄, x₅ be the observations with mean m and standard deviations s.
Then variance of x₁, x₂, x₃, x₄and x₅is s².
∴ Variance of kx₁, kx₂, kx₃, kx_4,and kx₅has variance k²s².
∴ Standard deviation of kx₁, kx₂, kx₃, kx_4,and kx₅ is $\sqrt{k^{2} s^{2}}$
i.e |k|s
Hence, the correct answer is option C.

Question 30:

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is
(a) 0
(b) 1
(c) 1.5
(d) 2.5

Answer:

Let coefficient of variation of first distribution be CV₁ and coefficient of variation of second distribution be CV₂
Let mean be given by $x_{1} and x_{2}$ respectively.

$∴ C V_{1} = \frac{σ_{1}}{x_{1}} \times 100 i . e 50 = \frac{σ_{1}}{30} \times 100 i . e σ_{1} = 15 and C V_{2} = \frac{σ_{2}}{x_{^{2}}} \times 100 i . e 60 = \frac{σ_{2}}{25} \times 100 i . e σ_{2} = 15 ∴ σ_{1} - σ_{2} = 15 - 15 = 0$

Hence, the correct answer is option A.

Question 31:

Given that $\frac{C}{5} = \frac{F - 32}{9}$ is the relation between temperature in ^∘C and ^∘F. The standard deviation of some temperature data in ^∘C is 5. If the data were converted in ^∘F, the variance would be
(a) 81
(b) 57
(c) 36
(d) 25

Answer:

Given standard deviation of same temperature in ^∘C is 5.
Let $σ_{1} = 5 for \frac{C}{5} = \frac{F - 32}{9}$
Given F = $\frac{9 C}{5} + 32$
Then standard deviation for F = $\frac{9}{5} σ_{1} + 0 (∵ S . D . of (x + k) = S . D of x)$
S.D for F = $\frac{9}{5} \times 5$
S.D for F = 9
∴ Variance for F = (standard deviation of F)² = 81

Hence, the correct answer is option A.

Question 32:

When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean deviations (in hours) from their mean is
(a) 178
(b) 179
(c) 220
(d) 356

Answer:

check ans
for 5 bulbs,
The lives observed are 1357, 1090, 1666, 1494, and 1623.
Mean of these observations $(say x) = \frac{1357 + 1090 + 1666 + 1494 + 1623}{5}$
i.e $x$ = 1446

observations	*deviation d_i* = $\|x_{i} - x\|$**
1357 1090 1666 1494 1623	89 356 220 48 177
Total	890

Hence mean deviation is

\frac{\sum d_{i}}{n} = \frac{890}{5} = 178

Hence, the correct answer is option A.

Question 33:

The following are the marks obtained by 9 students in Mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59. The mean deviation from the median is
(a) 9
(b) 10.5
(c) 12.67
(d) 14.67

Answer:

Given marks of 9 students in Mathematics test 50, 69, 20, 33, 53, 39, 40, 65, 59
Arranging the observations in ascending order 20, 33, 39, 40, 50, 53, 59, 65, 69

Since n is odd,
∴ median is ${(\frac{n + 1}{2})}^{th}$ term
i.e 50 is the median.
Calculation of mean deviation

x_i	$\|d_{i}\| = \|x_{i} - 50\|$
20 33 39 40 50 53 59 65 69	30 17 11 10 0 3 9 15 19
	$Σ d_{i} = 114$

∴ Mean deviation is \frac{1}{n} Σ d_{i} = \frac{114}{9} = 12.67

Hence, the correct answer is option C.

Question 1:

If $X$ is the mean of n values x₁, x₂, ......,x_n of variable X, then $\sum_{i = 1}^{n} (x_{i} - X)$ is always equal to _____________.

Answer:

$Let X is the mean of x_{1}, x_{2,} . . . . . x_{n} for variable X . then \sum_{i = 1}^{n} (x_{i} - X) = \sum_{i = 1}^{n} x_{i} - \sum_{i = 1}^{n} X = \sum_{i = 1}^{n} x_{i} - X \sum_{i = 1}^{n} 1 = \sum_{i = 1}^{n} x_{i} - n X \begin{array}{l} Since X = \frac{\sum_{i = 1}^{n} x_{i}}{n} \\ i . e \sum_{i = 1}^{n} x_{i} = n X \end{array} = n X - n X = 0 ∴ \sum_{i = 1}^{n} (x_{i} - X) = zero$

Question 2:

The mean deviation of the data is ________________ when measured from the median.

Answer:

Since out of mean, median
Mean deviation from median is least because of more accuracy.
∴ The mean deviation of the data is least when measured from the median.

Question 3:

The sum of the squares of the deviations of the values of the variable is ___________ when taken about their arithmetic mean.

Answer:

The sum of the squares of the deviations of the values of the variable is less when taken about their arithmetic mean.

Question 4:

If $X$ is the mean of n variables x₁, x₂,......., x_n of variable X and a has any value other than $X$ , is _____________ than $\sum_{i = 1}^{n} {(x_{i} - a)}^{2} .$

Answer:

check Quest 1
Let $X$ is the mean of n variables x₁, x₂,......., x_n of variable X.
Given a has any value other than $X$

Question 5:

The standard deviation is _____________ to the mean deviation taken from the arithmetic mean.

Answer:

The standard deviation is greater than or equal to the mean deviation taken from the arithmetic mean.

Since standard deviation deals with square of mean deviations about arithmetic mean.

Question 6:

If the variance of data is 121 then the standard deviation of the data is ______________.

Answer:

Since Standard deviation = $\sqrt{Variance}$

⇒ Standard deviation = $\sqrt{121}$

i.e Standard deviation = 11

Question 7:

The standard deviation of data is _________________ of any change in origin, but is _____________ on the change of scale.

Answer:

Standard deviation of any data is independent of any change in origin
i.e deviation remain same about any origin.
But it depends on change of scale.

Question 8:

Coefficient of variation = $\frac{A}{Mean}$ × 100, then A ________________.

Answer:

Since coefficient of variation is $\frac{σ}{\bar{X}} \times 100$
where $σ$ is Standard deviation
$∴ A = σ$
i.e A is standard deviation.

Question 9:

A set of n values x₁, x₂, .......x_n has standard deviation σ. The standard deviation of n values ax₁, ax₂, ...., a_xn is ________________.

Answer:

For x₁, x₂, ........x_n, Standard deviation is $σ$ (given)
Then variance of x₁, x₂, ..........x_n is $σ^{2}$
For ax₁, ax₂, ...........ax_n,
Variance is given by a²Var(x₁, x₂, ........x_n)²
i.e a² $σ^{2}$ .
$∴$ Standard deviation of ax₁, ax₂, ...........ax_{n

$= \sqrt{Variance (a x_{1}, a x_{2}, . . . . . . a x_{n})} = \sqrt{a^{2} σ^{2}}$}
Standard deviation of ax₁, ax₂, .......ax_n= |a|σ

Question 10:

If the variance of the data 2, 4, 5, 6, 8, 17 is 23.33, then the variance of 4, 8, 10, 12, 13, 34 is _____________________.

Answer:

Given ; Variance of 2, 4, 5, 6, 8, 17 is 23.33 = var(X)(say)
Using result ;
If variance of x₁, x₂,........, x_n is var(X)
Then variance of ax₁, ax₂,........., ax_n is a² var(X)
Then variance of 4, 8, 10, 12, 16, 32
i.e 2(2), 2(4), 2(5), 2(6), 2(8), 2(17)
is (2)² var(X)
= 2² × 23.33
= 4 × 23.33 = 93.32

Question 11:

The variance of first n natural numbers is ____________.

Answer:

For first n natural numbers. 1, 2, 3, ...........n

$\begin{array}{rcl} Mean & = & \frac{1 + 2 + 3 + . . . . . . . . + n}{n} \\ = & \frac{n (n + 1)}{2 n} \\ Mean & = & \frac{n + 1}{2} \end{array}$
$\begin{array}{rcl} Hence, Variance is σ^{2} & = & \frac{\sum_{} {(x_{i})}^{2}}{n} - {(mean)}^{2} \\ = & \frac{1^{2} + 2^{2} + 3^{2} + . . . . n^{2}}{n} - {(\frac{n + 1}{2})}^{2} \\ = & \frac{n (n + 1) (2 n + 1)}{6 n} - {(\frac{n + 1}{2})}^{2} \\ = & \frac{n + 1}{2} [\frac{(2 n + 1)}{3} - \frac{n + 1}{2}] \\ = & \frac{n + 1}{2} [\frac{4 n + 2 - 3 n - 3}{6}] \end{array}$
$Variance of first n natural numbers = \frac{n + 1}{2} (\frac{n - 1}{6}) = \frac{n^{2} - 1}{12} .$

Question 12:

If the standard deviation of a variable X is σ, then the S.D. of the variable $\frac{a X + b}{c}$ , where a, b, c are constants,is __________________.

Answer:

Let standard deviation of X be σ.
Then standard deviation of $(\frac{a X + b}{c})$
$= S . D (\frac{a}{c} (X) + (\frac{b}{c})) = |\frac{a}{c}| S . D (X) + S . D (\frac{b}{c}) = |\frac{a}{c}| σ + 0 i . e standard deviation of \frac{a X + b}{c} = |\frac{a}{c}| σ .$

Question 13:

If x₁, x₂,........x₁₈ are 18 observations such that $\sum_{i = 1}^{18} (x_{i} - 8) = 9 and \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45,$ then the standard deviation of these observations is _________________.

Answer:

Given for 18 observations,
Given $\sum_{i = 1}^{18} (x_{i} - 8) = 9, \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45$
$\begin{array}{rcl} Since variance of X & = & \frac{1}{n} (\sum_{i = 1}^{18} {(x_{i} - 8)}^{2}) - {(\sum_{i = 1}^{18} \frac{(x_{i} - 8)}{n})}^{2} \\ = & \frac{1}{18} (45) - 1 {(\frac{9}{18})}^{2} \\ = & \frac{5}{2} - {(\frac{1}{2})}^{2} = \frac{5}{2} - \frac{1}{4} = \frac{10 - 1}{4} \\ = & \frac{9}{4} \end{array}$
∴ Standard deviation = $\sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5$

Question 14:

If the standard deviation of first n natural numbers is 2, then n = ________________.

Answer:

Since standard deviation of first n natural number = $\sqrt{\frac{n^{2} - 1}{12}}$
Given standard deviation of first n natural number = 2

$i . e 2 = \sqrt{\frac{n^{2} - 1}{12}} i . e 4 = \frac{n^{2} - 1}{12} i . e 48 = n^{2} - 1 i . e n^{2} = 49 i . e n = 7$

Question 15:

If the variance of 1, 2, 3, 4, 5, ...,10 is $\frac{33}{4}$ , then the standard deviation of 3, 6, 9, 12, ...33 is ______________.

Answer:

Since Variance of 1, 2, 3, 4, 5, .......10 is $\frac{33}{4}$ .
Using if Variance of x₁, x₂, .......x_n is $σ$ then Variance of ax₁, ax₂, .......ax_n is a² $σ$ ²
$∴$ Variance of 3, 6, 9, .......30
i.e Variance of 3(1), 3(2), 3(3), .......3(10) is (3)²Variance (1, 2, 3, ......10)

$i . e 9 \times \frac{33}{4} = \frac{9 \times 33}{4}$
$∴ Standard deviation of 3, 6, 9, . . . . . . . 30 is \sqrt{\frac{9 \times 33}{4}} = \frac{3}{2} \sqrt{33}$

Question 1:

Write the variance of first n natural numbers.

Answer:

Sum of first n natural numbers $= \frac{n (n + 1)}{2}$

Mean, $\bar{X} = \frac{Sum of all the observations}{Total number of observations}$
$= \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2}$

$∴ σ^{2} = \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = \frac{\sum {(x_{i} - \frac{n + 1}{2})}^{2}}{n} = \frac{1}{n} \sum [{x_{i}}^{2} - x_{i} (n + 1) + {(\frac{n + 1}{2})}^{2}] = \frac{n (n + 1) (2 n + 1)}{6 n} - [\frac{n (n + 1)}{2}] (\frac{n + 1}{n}) + \frac{{(n + 1)}^{2}}{4 n} \times n = \frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{2} + \frac{{(n + 1)}^{2}}{4} = \frac{(n + 1) (n - 1)}{12} = \frac{(n^{2} - 1)}{12}$

Question 2:

If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.

Answer:

The sum of the squares of deviations for 10 observations, taken from their mean, is 2.5.
Square of each deviation = $\frac{2.5}{10} = 0.25$
Standard deviation = $\sqrt{0.25} = 0.5$

Question 3:

If x₁, x₂, ..., x_n are n values of a variable X and y₁, y₂, ..., y_n are n values of variable Y such that y_i = ax_i + b; i = 1, 2, ..., n, then write Var(Y) in terms of Var(X).

Answer:

$V a r (X) = \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} V a r (Y) = \frac{\sum {(y_{i} - \bar{Y})}^{2}}{n}$

We have:
$y_{i} = a x_{i} + b \bar{y} = \frac{\sum y_{i}}{n} = \frac{a \sum x_{i} + n b}{n} = a \bar{X} + b$
$Var (Y) = \frac{\sum_{} {(a x_{i} + b - a \bar{X} - b)}^{2}}{n} = \frac{\sum {(a x_{i} - a \bar{X})}^{2}}{n} = a^{2} \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = a^{2} V a r (X)$

Question 4:

If X and Y are two variates connected by the relation $Y = \frac{a X + b}{c}$ and Var (X) = σ², then write the expression for the standard deviation of Y.

Answer:

$Y = \frac{a X + b}{c} Y = \frac{\sum y_{i}}{n} = \frac{\frac{a \sum X + n b}{c}}{n} = \frac{a \sum X}{n c} + \frac{n b}{n c} = \frac{a \bar{X}}{c} + \frac{b}{c}$

We know:

$Var (X) = \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = σ^{2} Var (Y) = \frac{\sum {(y_{i} - \bar{Y})}^{2}}{n} = \frac{\sum {(\frac{a X}{c} + \frac{b}{c} - \frac{a}{c} \bar{X} - \frac{b}{c})}^{2}}{n} = \frac{\sum {(\frac{a X}{c} - \frac{a}{c} \bar{X})}^{2}}{n} = {(\frac{a}{c})}^{2} \frac{\sum {(x_{i} - \bar{X})}^{2}}{n} = {(\frac{a}{c})}^{2} σ^{2} SD (σ) = \sqrt{{(\frac{a}{c})}^{2} σ^{2}} = |\frac{a}{c}| σ$

Question 5:

In a series of 20 observations, 10 observations are each equal to k and each of the remaining half is equal to − k. If the standard deviation of the observations is 2, then write the value of k.

Answer:

$n = 20 d_{i} = x_{i} - a = x_{i} - \frac{\sum x_{i}}{20} = x_{i} - 0 = x_{i} \Rightarrow \sum d_{i} = \sum x_{i} = 0 \Rightarrow \sum {d_{i}}^{2} = 20 k^{2} \Rightarrow σ^{2} = \frac{\sum {d_{i}}^{2}}{n} - {(\frac{\sum d_{i}}{n})}^{2} = \frac{20 k^{2}}{20} - 0 = k^{2} \Rightarrow σ = 2 = \sqrt{k^{2}} \Rightarrow k = \pm 2$

Question 6:

If each observation of a raw data whose standard deviation is σ is multiplied by a, then write the S.D. of the new set of observations.

Answer:

Standard deviation, $σ = \sqrt{\frac{\sum_{i} {(x_{i} - x)}^{2}}{n}}$

Here, $x$ represents the arithmetic mean.

Multiplying each $x_{i}$ by $a$ :
$x_{new} = \underset{i}{\frac{1}{n} \sum} a . x_{i} = a \times \frac{1}{n} \sum_{i} x_{i} = a . x_{old}$

$New standard deviation, σ_{new} = \sqrt{\frac{\sum_{i} {(a . x_{i} - a . x)}^{2}}{n}} = \sqrt{\frac{\sum_{i} a^{2} . {(x_{i} - x)}^{2}}{n}} = |a| \sqrt{\frac{\sum_{i} {(x_{i} - x)}^{2}}{n}} = |a| . σ$

Question 7:

If a variable X takes values 0, 1, 2,..., n with frequencies ⁿC₀, ⁿC₁, ⁿC₂ , ... , ⁿC_n, then write variance X.

Answer:

$x = \frac{\sum_{i = 0}^{n} x_{i} f_{i}}{\sum_{i = 0}^{n} f_{i}} = \frac{0 \times {}^{n}C_{o} + 1 \times {}^{n}C_{1} + . . . + n \times {}^{n}C_{n}}{{}^{n}C_{o} + {}^{n}C_{1} + . . . + {}^{n}C_{n}} \Rightarrow x = \frac{n \times 2^{n - 1}}{\frac{2^{n}}{n + 1}} = \frac{n (n + 1)}{2} ∴ V a r (X) = σ^{2} = \frac{1}{n} \sum_{i = 0}^{n} {(x_{i} - x)}^{2} = \frac{1}{n} {[(0 + 1 + 2 + . . . . + n) - n x]}^{2} \Rightarrow σ^{2} = \frac{1}{n} {[\frac{n (n + 1)}{2} - \frac{n \times n (n + 1)}{2}]}^{2} = \frac{1}{n} {[\frac{n (n + 1)}{2} (1 - n)]}^{2} = \frac{n^{2}}{4 n} {(n + 1)}^{2} {(n - 1)}^{2}$

Question 1:

Calculate the mean deviation about the median of the following observations:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Answer:

Formula used for mean deviation:
$M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}| H e r e, d_{i} = x_{i} - M$
M = Median

i) Arranging the data in ascending order:
2354, 2780, 3011, 3020, 3541, 4150, 5000

Here, median $(M) = 3020$ and n = 7.

$x_{i}$	$\|d_{i}\|$ = $\|x_{i} - 3020\|$
3011	9
2780	240
3020	0
2354	666
3541	521
4150	1130
5000	1980
Total	4546

M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|

\Rightarrow M D = \frac{1}{7} \times 4546 = 649.42

ii) Arranging the data in ascending order:
34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here, n is equal to 10.
Median is the arithmetic mean of the fifth and the sixth observation.

M e d i a n, M = \frac{46 + 48}{2} = 47

x_i	$\|d_{i}\| = \|x_{i} - M\|$
38	9
70	23
48	1
34	13
42	5
55	8
63	16
46	1
54	7
44	3
Total	86

M D = \frac{1}{10} \times 86 = 8.6

iii) Arranging the data in ascending order:
30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here,

n = 10

Also, median is the AM of the fifth and the sixth observation.

M e d i a n, M = \frac{42 + 44}{2} = 43

x_i	$\|d_{i}\| = \|x_{i} - M\|$
34	9
66	23
30	13
38	5
44	1
50	7
40	3
60	17
42	1
51	8
Total	87

M D = \frac{1}{10} \times 87 = 8.7

iv) Arranging the data in ascending order.
22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here, n = 10 .

Also, median is the AM of the fifth and the sixth observation.

M e d i a n, M = \frac{28 + 29}{2} = 28.5

x_i	$\|d_{i}\| = \|x_{i} - M\|$
22	6.5
24	4.5
30	1.5
27	1.5
29	0.5
31	2.5
25	3.5
28	0.5
41	12.5
41	13.5
Total	47

M D = \frac{1}{10} \times 47 = 4.7

v) Arranging the data in ascending order:
34, 38, 42, 44, 47, 48, 53, 55, 63, 70

Here,

n = 10

.
Also, median is the AM of the fifth and the sixth observation.

M e d i a n, M = \frac{47 + 48}{2} = 47.5

x_i	$\|d_{i}\| = \|x_{i} - M\|$
38	9.5
70	22.5
48	0.5
34	13.5
63	15.5
42	5.5
55	7.5
44	3.5
53	5.5
47	0.5
Total	84

M D = \frac{1}{10} \times 84 = 8.4

Question 2:

Calculate the mean deviation from the mean for the following data:
(i) 4, 7, 8, 9, 10, 12, 13, 17
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer:

Formula used for finding the mean deviation about the mean is given below:

$M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|, w h e r e |d_{i}| = |x_{i} - x|$

i)
Let $x$ be the mean of the given data.

$x = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} = 10$

$x_{i}$	$\|d_{i}\| = \|x_{i} - \bar{x}\|$
4	6
7	3
8	2
9	1
10	0
12	2
13	3
17	7
Total	24

M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|

M D = \frac{1}{8} \times 24 = 3

ii)
Let

x

be the mean of the given data.

x = \frac{13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17}{12} = 14

$x_{i}$	$\|d_{i}\| = \|x_{i} - \bar{x}\|$
13	1
17	3
16	2
14	0
11	3
13	1
10	4
16	2
11	3
18	4
12	2
17	3
Total	28

M D = \frac{1}{12} \times 28 = 2.33

iii)
Let

x

be the mean of the given data.

x = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10} = 50

$x_{i}$	$\|d_{i}\| = \|x_{i} - \bar{x}\|$
38	12
70	20
48	2
40	10
42	8
55	5
63	13
46	4
54	4
44	6
Total	84

M D = \frac{1}{10} \times 84 = 8.4

iv)
Let

x

be the mean of the given data.

x = \frac{36 + 746 + 42 + 60 + 45 + 53 + 46 + 51 + 59}{10} = 50

$x_{i}$	$\|d_{i}\| = \|x_{i} - \bar{x}\|$
36	14
72	22
46	4
42	8
60	10
45	5
53	3
46	4
51	1
49	1
Total	72

M D = \frac{1}{10} \times 72 = 7.2

Question 3:

Calculate the mean deviation of the following income groups of five and seven members from their medians:

I Income in Rs.	II Income in Rs.
4000 4200 4400 4600 4800	300 4000 4200 4400 4600 4800 5800

Answer:

Calculate the mean deviation for the first data set.
The data is already arranged in ascending order.
For this data set, n is equal to 5.
Also, median, $M = 4400$
$M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|, w h e r e |d_{i}| = |x_{i} - M|$

$x_{i}$	$\|d_{i}\| = \|x_{i} - M\|$
4000	400
4200	200
4400	0
4600	200
4800	400
Total	1200

M D = \frac{1}{5} \times 1200 = 240

Therefore, for the income of families in the first group, the mean deviation from the median is Rs 240.

Now, consider the second data set. This is also arranged in ascending order.
Here,

n = 7

.
Also, median,

M = 4400

x_i	$\|d_{i}\| = \|x_{i} - M\|$
300	4100
4000	400
4200	200
4400	0
4600	200
4800	400
5800	1400
Total	6700

M D = \frac{1}{7} \times 6700 = 957.14

Therefore, for the income of families in the second group, the mean deviation from the median is Rs 957.14.

Question 4:

The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find mean deviation from median
(ii) Find mean deviation from the mean also.

Answer:

i) Formula for the mean deviation from the median is as follows:
$M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|, w h e r e |d_{i}| = |x_{i} - M|$

Arranging the data in ascending order for finding the median:
15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79

Here, $n = 10$ .
Therefore, median is the average of the fifth and the sixth observations.

$M = \frac{40 + 52.3}{2} = 46.15$

$x_{i}$	$\|d_{i}\| = \|x_{i} - 46.15\|$
40	6.15
52.3	6.15
55.2	9.05
72.9	26.75
52.8	6.65
79	32.85
32.5	13.65
15.2	30.95
27.9	18.25
32	14.15
Total	164.6

M D = \frac{1}{10} \times 164.6 = 16.46

Mean deviation from median in 16.4 cm.

ii)
Let

\bar{x}

be the mean of the given data set.

x = \frac{40 + 52.3 + 55.2 + 72.9 + 52.8 + 79 + 32.5 + 15.2 + 27.9 + 30.2}{10} = 45.98

$x_{i}$	$\|d_{i}\| = \|x_{i} - 45.98\|$
40	5.98
52.3	6.32
55.2	9.22
72.9	26.92
52.8	6.82
79	33.02
32.5	13.48
15.2	30.78
27.9	18.08
32	13.98
Total	164.6

M D = \frac{1}{10} \times 164.6 = 16.46

Mean deviation from the mean is 16.4 cm.

Question 5:

In question 1 (iii), (iv), (v) find the number of observations lying between $X$ − M.D. and $X$ + M.D, where M.D. is the mean deviation from the mean.

Answer:

iii)
Let $\bar{x}$ be the mean of the data set.

$x = \frac{34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51}{10} = 45.5$

$M D = \frac{1}{n} \sum_{i = 1}^{n} |d_{i}|, w h e r e |d_{i}| = |x_{i} - x|$

$x_{i}$	$\|d_{i}\| = \|x_{i} - 45.5\|$
34	11.5
66	20.5
30	15.5
38	7.5
44	1.5
50	4.5
40	5.5
60	14.5
42	3.5
51	5.5
Total	90

M D = \frac{1}{10} \times 90 = 9

x - M . D . = 45.5 - 9 = 36.5 Also, x + M . D . = 45.5 + 9 = 54.5

Hence, there are 6 observations between 36.5 and 54.5.

iv)
Let

\bar{x}

be the mean of the data set.

x = \frac{22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42}{10} = 29.9

$x_{i}$	$\|d_{i}\| = \|x_{i} - 29.9\|$
22	7.9
24	5.9
30	0.1
27	2.9
29	0.9
31	1.1
25	4.9
28	1.9
41	11.9
42	12.1
Total	48.8

M D = \frac{1}{10} \times 48.8 = 4.88

x - M . D . = 29.9 - 4.88 = 25.02, a n d x + M . D . = 29.9 + 4.88 = 34.78

There are 5 observations between 25.02 and 34.78.

v)
Let

\bar{x}

be the mean of the data set.

x = \frac{38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47}{10} = 49.4

$x_{i}$	$\|d_{i}\| = \|x_{i} - 49.4\|$
38	11.4
70	20.6
48	1.4
34	15.4
63	13.6
42	7.4
55	5.6
44	5.4
53	3.6
47	2.4
Total	86.8

M D = \frac{1}{10} \times 86.6 = 8.68

x - M D = 49.4 - 8.68 = 40.72 a n d x + M D = 49.4 + 8.68 = 58.08

There are 6 observations between 40.72 and 58.08.

Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 32 - Statistics

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