NCERT Solutions for Class 11 Science Math Chapter 15 - Statistics

Answer:

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e., are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

Answer:

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

Answer:

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e.are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations,, are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

Answer:

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e.are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,, are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

Answer:

xi	fi	fi xi
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
	25	350		158

Answer:

xi	fi	fi xi
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
	80	4000		1280

Answer:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi	fi	c.f.
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26

Here, N = 26, which is even.

Median is the mean of 13^th and 14^th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e.are

|xi – M|	2	0	2	3	5	8
fi	8	6	2	2	2	6
fi |xi – M|	16	0	4	6	10	48

and

Answer:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi	fi	c.f.
15	3	3
21	5	8
27	6	14
30	7	21
35	8	29

Here, N = 29, which is odd.

observation = 15^th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e.are

|xi – M|	15	9	3	0	5
fi	3	5	6	7	8
fi |xi – M|	45	45	18	0	40

∴

Answer:

The following table is formed.

Income per day	Number of persons fi	Mid-point xi	fi xi
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
	50		17900		7896

Here,

Answer:

The following table is formed.

Height in cms	Number of boys fi	Mid-point xi	fi xi
95-105	9	100	900	25.3	227.7
105-115	13	110	1430	15.3	198.9
115-125	26	120	3120	5.3	137.8
125-135	30	130	3900	4.7	141
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247

Here,

Answer:

The following table is formed.

Marks	Number of girls fi	Cumulative frequency (c.f.)	Mid-point xi	|xi – Med.|	fi |xi – Med.|
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
20-30	14	28	25	2.85	39.9
30-40	16	44	35	7.15	114.4
40-50	4	48	45	17.15	68.6
50-60	2	50	55	27.15	54.3
	50				517.1

The class interval containing theor 25^th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

Here, l = 20, C = 14, f = 14, h = 10, and N = 50

∴ Median =

Thus, mean deviation about the median is given by,

Answer:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age	Number fi	Cumulative frequency (c.f.)	Mid-point xi	|xi – Med.|	fi |xi – Med.|
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	160
50.5-55.5	9	100	53	15	135
	100				735

The class interval containing theor 50^th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,

Answer:

6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

xi
6	–3	9
7	–2	4
10	–1	1
12	3	9
13	4	16
4	–5	25
8	–1	1
12	3	9
		74

Answer:

The mean of first n natural numbers is calculated as follows.

Answer:

The first 10 multiples of 3 are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, n = 10

The following table is obtained.

xi
3	–13.5	182.25
6	–10.5	110.25
9	–7.5	56.25
12	–4.5	20.25
15	–1.5	2.25
18	1.5	2.25
21	4.5	20.25
24	7.5	56.25
27	10.5	110.25
30	13.5	182.25
		742.5

Answer:

The data is obtained in tabular form as follows.

xi	f i	fixi
6	2	12	–13	169	338
10	4	40	–9	81	324
14	7	98	–5	25	175
18	12	216	–1	1	12
24	8	192	5	25	200
28	4	112	9	81	324
30	3	90	11	121	363
	40	760			1736

Here, N = 40,

Answer:

The data is obtained in tabular form as follows.

xi	f i	fixi
92	3	276	–8	64	192
93	2	186	–7	49	98
97	3	291	–3	9	27
98	2	196	–2	4	8
102	6	612	2	4	24
104	3	312	4	16	48
109	3	327	9	81	243
	22	2200			640

Here, N = 22,

Answer:

The data is obtained in tabular form as follows.

xi	fi		yi2	fiyi	fiyi2
60	2	–4	16	–8	32
61	1	–3	9	–3	9
62	12	–2	4	–24	48
63	29	–1	1	–29	29
64	25	0	0	0	0
65	12	1	1	12	12
66	10	2	4	20	40
67	4	3	9	12	36
68	5	4	16	20	80
	100	220		0	286

Mean,

Answer:

Class	Frequency fi	Mid-point xi		yi2	fiyi	fiyi2
0-30	2	15	–3	9	–6	18
30-60	3	45	–2	4	–6	12
60-90	5	75	–1	1	–5	5
90-120	10	105	0	0	0	0
120-150	3	135	1	1	3	3
150-180	5	165	2	4	10	20
180-210	2	195	3	9	6	18
	30				2	76

Mean,

Answer:

Mean,

Answer:

Mean,

Answer:

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Mean,

Answer:

Firstly, the standard deviation of group A is calculated as follows.

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

Answer:

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

Answer:

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ₁) =

Variance of the distribution of wages in firm = 121

∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

Answer:

The mean and the standard deviation of goals scored by team A are calculated as follows.

Thus, the mean of both the teams is same.

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

Answer:

Here, N = 50

∴ Mean,

Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

Answer:

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

From (1), we obtain

x² + y² + 2xy = 144 …(3)

From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 80 – 64 = 16

⇒ x – y = ± 4 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = –4

Thus, the remaining observations are 4 and 8.

Answer:

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

From (1), we obtain

x² + y² + 2xy = 196 … (3)

From (2) and (3), we obtain

2xy = 196 – 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 100 – 96

⇒ (x – y)² = 4

⇒ x – y = ± 2 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.

Answer:

Let the observations be x₁, x₂, x₃, x₄, x₅, and x₆.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y_i, then

From (1) and (2), it can be observed that,

Substituting the values of x_i and in (2), we obtain

Therefore, variance of new observations =

Hence, the standard deviation of new observations is

Answer:

The given n observations are x₁, x₂ … x_n.

Mean =

Variance = σ²

If each observation is multiplied by a and the new observations are y_i, then

Therefore, mean of the observations, ax₁, ax₂ … ax_n, is .

Substituting the values of x_iand in (1), we obtain

Thus, the variance of the observations, ax₁, ax₂ … ax_n, is a² σ².

Answer:

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean

$$\begin{gathered} \mathrm{Standard} \mathrm{deviation}, \mathrm{\sigma } = \sqrt{\frac{1}{n}\sum _{i=1}^n{x_i}^2 - \frac{1}{n^2}{\left(\sum _{i=1}^n{x}_i\right)}^2} \\ \Rightarrow 2 = \sqrt{\frac{1}{n}\sum _{i=1}^n{x_i}^2 - {\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^2} \\ \Rightarrow 2 = \sqrt{\frac{1}{n}\sum _{i=1}^n{x_i}^2 - {\left(\overline{x}\right)}^2} \left[\mathrm{as}, \frac{1}{n}\sum _{i=1}^{n}x = \overline{x} \right] \\ \Rightarrow 2 = \sqrt{\frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^n{x_i}^2 - {\left(10\right)}^2} \\ \Rightarrow 4 = \frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^n{x_i}^2 - 100 \\ \Rightarrow \frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^n{x_i}^2 = 104 \\ \Rightarrow \mathrm{Incorrect}\sum _{i=1}^n{x_i}^2 = 2080 \\ \mathrm{Now}, \mathrm{correct}\sum _{i=1}^n{x_i}^2 = \mathrm{Incorrect}\sum _{i=1}^n{x_i}^2 - {\left(8\right)}^2 \\ \Rightarrow \mathrm{correct}\sum _{i=1}^n{x_i}^2 = 2080 - 64 = 2016 \\ \therefore \mathrm{correct} \mathrm{Standard} \mathrm{Deviation} = \sqrt{\frac{1}{n}\mathrm{correct}\sum _{i=1}^n{x_i}^2 - {\left(\mathrm{correct} \mathrm{mean}\right)}^2} \\ \Rightarrow \mathrm{correct} \mathrm{Standard} \mathrm{Deviation} = \sqrt{\frac{1}{19}\times 2016 - {\left(\frac{192}{19}\right)}^2} \\ \Rightarrow \mathrm{correct} \mathrm{Standard} \mathrm{Deviation} = \sqrt{\frac{2016}{19}-{\left(\frac{192}{19}\right)}^2} \\ \Rightarrow \mathrm{correct} \mathrm{Standard} \mathrm{Deviation} = \frac{\sqrt{1440}}{19} = \frac{12\sqrt{10}}{19} \\ \Rightarrow \mathrm{correct} \mathrm{Standard} \mathrm{Deviation} = \frac{12 \times 3.162}{19} = 1.997 \end{gathered}$$

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

Answer:

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by .

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Answer:

Number of observations (n) = 100

Incorrect mean

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940