Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 18 - Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

Answer:

Mean of given observations = $\frac{Sum of given observations}{Total number of observations}$

$∴ 11 = \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} \Rightarrow 55 = 5 x + 20 \Rightarrow 5 x = 55 - 20 \Rightarrow 5 x = 35 \Rightarrow x = \frac{35}{5} \Rightarrow x = 7$

Hence, the value of x is 7.

Answer:

Mean of given observations = $\frac{Sum of given observations}{Total number of observations}$

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500

Then, new mean = $\frac{500}{25} = 20$

Thus, the new mean will be 20.

Answer:

Here, h = 20

Let the assumed mean, A be 60.

Class	*Mid-Values(x_i)*	*Frequency (f_i)*	$d_{i} = x_{i} - 60$	$u_{i} = \frac{x_{i} - 60}{20}$	$f_{i} u_{i}$
10−30	20	15	−40	−2	−30
30−50	40	18	−20	−1	−18
50−70	60	25	0	0	0
70−90	80	10	20	1	10
90−110	100	2	40	2	4
		$\underset{}{\sum f_{i} = 70}$			$\underset{}{\sum f_{i} u_{i} = - 34}$

We know

Mean $= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})$

∴ Mean of the given frequency distribution

$= 60 + 20 \times (\frac{- 34}{70}) = 60 - 9.71 = 50.29 (Approx)$

Thus, the mean of the given frequency distribution is approximately 50.29.

Answer:

Here, h = 20

Let the assumed mean, A be 70.

Class	*Mid-Values(x_i)*	*Frequency (f_i)*	$d_{i} = x_{i} - 70$	$u_{i} = \frac{x_{i} - 70}{20}$	$f_{i} u_{i}$
0−20	10	6	−60	−3	−18
20−40	30	8	−40	−2	−16
40−60	50	10	−20	−1	−10
60−80	70	12	0	0	0
80−100	90	6	20	1	6
100−120	110	5	40	2	10
120−140	130	3	60	3	9
		$\underset{}{\sum f_{i} = 50}$			$\underset{}{\sum f_{i} u_{i} = - 19}$

We know

Mean $= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})$

∴ Mean of the given frequency distribution

$= 70 + 20 \times (\frac{- 19}{50}) = 70 - 7.6 = 62.4$

Thus, the mean of the given frequency distribution is 62.4.

Answer:

Here, h = 6

Let the assumed mean, A be 21.

Number of days	*Mid-Values(x_i)*	*Number of Students (f_i)*	$d_{i} = x_{i} - 21$	$u_{i} = \frac{x_{i} - 21}{6}$	$f_{i} u_{i}$
0−6	3	10	−18	−3	−30
6−12	9	11	−12	−2	−22
12−18	15	7	−6	−1	−7
18−24	21	4	0	0	0
24−30	27	4	6	1	4
30−36	33	3	12	2	6
36−42	39	1	18	3	3
		$\underset{}{\sum f_{i} = 40}$			$\underset{}{\sum f_{i} u_{i} = - 46}$

We know

Mean $= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})$

∴ Mean number of days a student was absent

$= 21 + 6 \times (\frac{- 46}{40}) = 21 - 6.9 = 14.1$

Thus, the mean number of days a student was absent is 14.1.

Answer:

Here, h = 200

Let the assumed mean, A be ₹1000.

Expenses (in ₹)	*Mid-Values(x_i)*	*Number of Families(f_i)*	$d_{i} = x_{i} - 1000$	$u_{i} = \frac{x_{i} - 1000}{200}$	$f_{i} u_{i}$
500−700	600	6	−400	−2	−12
700−900	800	8	−200	−1	−8
900−1100	1000	10	0	0	0
1100−1300	1200	9	200	1	9
1300−1500	1400	7	400	2	14
		$\underset{}{\sum f_{i} = 40}$			$\underset{}{\sum f_{i} u_{i} = 3}$

We know

Mean

= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})

∴ Mean daily expenses

$= 1000 + 200 \times (\frac{3}{40}) = 1000 + 15 = ₹ 1015$

Thus, the mean daily expenses is ₹1015.

Answer:

Here, h = 4

Let the assumed mean, A be 78.

Number of Heartbeats per Minute	*Mid-Values(x_i)*	*Number of Patients (f_i)*	$d_{i} = x_{i} - 78$	$u_{i} = \frac{x_{i} - 78}{4}$	$f_{i} u_{i}$
64−68	66	6	−12	−3	−18
68−72	70	8	−8	−2	−16
72−76	74	10	−4	−1	−10
76−80	78	12	0	0	0
80−84	82	3	4	1	3
84−88	86	1	8	2	2
		$\underset{}{\sum f_{i} = 40}$			$\underset{}{\sum f_{i} u_{i} = - 39}$

We know

Mean

= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})

∴ Mean heartbeats per minute for these patients

= 78 + 4 \times (\frac{- 39}{40}) = 78 - 3.9 = 74.1

Thus, the mean heartbeats per minute for these patients is 74.1.

Answer:

Here, h = 10

Let the assumed mean, A be 35.

Age (in years)	*Mid-Values(x_i)*	*Number of persons (f_i)*	$d_{i} = x_{i} - 35$	$u_{i} = \frac{x_{i} - 35}{10}$	$f_{i} u_{i}$
0−10	5	105	−30	−3	−315
10−20	15	222	−20	−2	−444
20−30	25	220	−10	−1	−220
30−40	35	138	0	0	0
40−50	45	102	10	1	102
50−60	55	113	20	2	226
60−70	65	100	30	3	300
		$\underset{}{\sum f_{i} = 1000}$			$\underset{}{\sum f_{i} u_{i} = - 351}$

We know

Mean $= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})$

∴ Mean age of the persons visiting the marketing centre on that day

$= 35 + 10 \times (\frac{- 351}{1000}) = 35 - 3.51 = 31.49$

Thus, the mean age of the persons visiting the marketing centre on that day 31.49 years.

Answer:

Class	*Mid-Values(x_i)*	*Frequency (f_i)*	*f_i x_i*
0−20	10	12	120
20−40	30	15	450
40−60	50	32	1600
60−80	70	x	70x
80−100	90	13	1170
		$\underset{}{\sum f_{i} = 72 + x}$	$\underset{}{\sum f_{i} x_{i} = 3340 + 70 x}$

Mean of the given frequency distribution = 53

We know

Mean

= \frac{\sum_{} f_{i} x_{i}}{\sum_{} f_{i}}

∴ \frac{3340 + 70 x}{72 + x} = 53 \Rightarrow 3340 + 70 x = 3816 + 53 x \Rightarrow 70 x - 53 x = 3816 - 3340

\Rightarrow 17 x = 476 \Rightarrow x = \frac{476}{17} = 28

Thus, the value of x is 28.

Answer:

The given data is shown as follows:

Daily pocket allowance (in ₹)	Number of children *(f_i)*	*Class mark (x_i)*	*f_ix_i*
11−13	7	12	84
13−15	6	14	84
15−17	9	16	144
17−19	13	18	234
19−21	f	20	20f
21−23	5	22	110
23−25	4	24	96
Total	∑ f_i = 44 + f		∑ f_ix_i = 752 + 20f

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 18 = \frac{752 + 20 f}{44 + f} \Rightarrow 18 (44 + f) = 752 + 20 f \Rightarrow 792 + 18 f = 752 + 20 f \Rightarrow 20 f - 18 f = 792 - 752 \Rightarrow 2 f = 40 \Rightarrow f = 20

Hence, the value of f is 20.

Answer:

The given data is shown as follows:

Class	Frequency *(f_i)*	*Class mark (x_i)*	*f_ix_i*
0−20	7	10	70
20−40	p	30	30p
40−60	10	50	500
60−80	9	70	630
80−100	13	90	1170
Total	∑ f_i = 39 + p		∑ f_ix_i = 2370 + 30p

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 54 = \frac{2370 + 30 p}{39 + p} \Rightarrow 54 (39 + p) = 2370 + 30 p \Rightarrow 2106 + 54 p = 2370 + 30 p \Rightarrow 54 p - 30 p = 2370 - 2106 \Rightarrow 24 p = 264 \Rightarrow p = 11

Thus, the value of p is 11.

Answer:

The given data is shown as follows:

Class interval	Frequency *(f_i)*	*Class mark (x_i)*	*f_ix_i*
0−10	7	5	35
10−20	10	15	150
20−30	x	25	25x
30−40	13	35	455
40−50	y	45	45y
50−60	10	55	550
60−70	14	65	910
70−80	9	75	675
Total	∑ f_i = 63 + x + y		∑ f_ix_i = 2775 + 25x + 45y

Sum of the frequencies = 100

\Rightarrow \sum_{i} f_{i} = 100 \Rightarrow 63 + x + y = 100 \Rightarrow x + y = 100 - 63 \Rightarrow x + y = 37 \Rightarrow y = 37 - x . . . . (1)

Now, The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 42 = \frac{2775 + 25 x + 45 y}{100} \Rightarrow 4200 = 2775 + 25 x + 45 y \Rightarrow 4200 - 2775 = 25 x + 45 y \Rightarrow 1425 = 25 x + 45 (37 - x) [from (1)] \Rightarrow 1425 = 25 x + 1665 - 45 x \Rightarrow 20 x = 1665 - 1425 \Rightarrow 20 x = 240 \Rightarrow x = 12

If x = 12, then y = 37 − 12 = 25

Thus, the value of x is 12 and y is 25.

Answer:

The given data is shown as follows:

Expenditure (in ₹)	Number of families *(f_i)*	*Class mark (x_i)*	*f_ix_i*
140−160	5	150	750
160−180	25	170	4250
180−200	f₁	190	190f₁
200−220	f₂	210	210f₂
220−240	5	230	1150
Total	∑ f_i = 35 + f₁ + f₂		∑ f_ix_i = 6150 + 190f₁ + 210f₂

Sum of the frequencies = 100

\Rightarrow \sum_{i} f_{i} = 100 \Rightarrow 35 + f_{1} + f_{2} = 100 \Rightarrow f_{1} + f_{2} = 100 - 35 \Rightarrow f_{1} + f_{2} = 65 \Rightarrow f_{2} = 65 - f_{1} . . . . (1)

Now, The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 188 = \frac{6150 + 190 f_{1} + 210 f_{2}}{100} \Rightarrow 18800 = 6150 + 190 f_{1} + 210 f_{2} \Rightarrow 18800 - 6150 = 190 f_{1} + 210 f_{2} \Rightarrow 12650 = 190 f_{1} + 210 (65 - f_{1}) [from (1)] \Rightarrow 12650 = 190 f_{1} - 210 f_{1} + 13650 \Rightarrow 20 f_{1} = 13650 - 12650 \Rightarrow 20 f_{1} = 1000 \Rightarrow f_{1} = 50

If f₁ = 50, then f₂ = 65 − 50 = 15

Thus, the value of f₁ is 50 and f₂ is 15.

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-20	7	10	70
20-40	f₁	30	30 f₁
40-60	12	50	600
60-80	18- f₁	70	1260-70 f₁
80-100	8	90	720
100-120	5	110	550
	$\sum f_{i} = 50$		$\sum (f_{i} \times x_{i}) = 3200 - 40 f_{1}$

$We have: 7+ f_{1} + 12 + f_{2} + 8 + 5 = 50 \Rightarrow f_{1} + f_{2} = 18 \Rightarrow f_{2} = 18 - f_{1} ∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} \Rightarrow 57 .6= \frac{3200 - 40 f_{1}}{50} [∵ Mean=57 .6] \Rightarrow 40 f_{1} = 320 ∴ f_{1} = 8 And f_{2} = 18 - 8 \Rightarrow f_{2} =10 ∴ The missing frequencies are f_{1} = 8 and f_{2} =10.$

Answer:

Here, h = 80

Let the assumed mean, A be 200.

Class	*Mid-Values(x_i)*	*Frequency(f_i)*	$d_{i} = x_{i} - 200$	$u_{i} = \frac{x_{i} - 200}{80}$	$f_{i} u_{i}$
0−80	40	20	−160	−2	−40
80−160	120	25	−80	−1	−25
160−240	200	x	0	0	0
240−320	280	y	80	1	y
320−400	360	10	160	2	20
		$\underset{}{\sum f_{i} = x + y + 55}$			$\underset{}{\sum f_{i} u_{i} = - 45 + y}$

\underset{}{\sum f_{i} = x + y + 55} = 100

(Given)

\Rightarrow x + y = 45 . . . . . (1)

We know

Mean

= A + h \times (\frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}})

∴ 200 + 80 \times (\frac{- 45 + y}{100}) = 188

(Given)

\Rightarrow \frac{- 45 + y}{100} = \frac{188 - 200}{80} = - 0.15 \Rightarrow - 45 + y = - 15 \Rightarrow y = 45 - 15 = 30

Substituting the value of y in (1), we get

x + 30 = 45

⇒ x = 45 − 30 = 15

Thus, the missing frequencies x and y are 15 and 30, respectively.

Answer:

Using Direct method, the given data is shown as follows:

Literacy rate (%)	Number of cities *(f_i)*	*Class mark (x_i)*	*f_ix_i*
45−55	4	50	200
55−65	11	60	660
65−75	12	70	840
75−85	9	80	720
85−95	4	90	360
Total	∑ f_i = 40		∑ f_ix_i = 2780

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{2780}{40} = 69.5

Thus, the mean literacy rate is 69.5%.

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 25)$	$(f_{i} \times d_{i})$
0-10	12	5	-20	-240
10-20	18	15	-10	-180
20-30	27	25=A	0	0
30-40	20	35	10	200
40-50	17	45	20	340
50-60	6	55	30	180
	$\sum f_{i} = 100$			$\sum (f_{i} \times d_{i}) = 300$

$Let A = 25 be the assumed mean . Then we have: Mean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} = 25+ \frac{300}{100} =28 ∴ \bar{x} = 28$

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 150)$	$(f_{i} \times d_{i})$
100-120	10	110	-40	-400
120-140	20	130	-20	-400
140-160	30	150=A	0	0
160-180	15	170	20	300
180-200	5	190	40	200
	$\sum f_{i} = 80$			$\sum (f_{i} \times d_{i}) = - 300$

$Let A = 150 be the assumed mean . Then we have: Mean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} = 150 - \frac{300}{80} =150 - 3 .75 ∴ \bar{x} = 146.25$

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 50)$	$(f_{i} \times d_{i})$
0-20	20	10	-40	-800
20-40	35	30	-20	-700
40-60	52	50=A	0	0
60-80	44	70	20	880
80-100	38	90	40	1520
100-120	31	110	60	1860
	$\sum f_{i} = 220$			$\sum (f_{i} \times d_{i}) = 2760$

$Let A = 50 be the assumed mean . Then we have: Now, m ean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} =50+ \frac{2760}{220} =50+12 .55 ∴ \bar{x} = 62.55$

Answer:

Let us choose a = 25, h = 10, then d_i = x_i − 25 and u_{i = $\frac{x_{i} - 25}{10}$}.

Using Step-deviation method, the given data is shown as follows:

Class	Frequency *(f_i)*	*Class mark (x_i)*	d_i = x_i − 25	u_{i = $\frac{x_{i} - 25}{10}$}	*f_iu_i*
0−10	7	5	−20	−2	−14
10−20	10	15	−10	−1	−10
20−30	15	25	0	0	0
30−40	8	35	10	1	8
40−50	10	45	20	2	20
Total	∑ f_i = 50				∑ f_iu_i = 4

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 25 + \frac{4}{50} \times 10 = 25 + \frac{4}{5} = \frac{125 + 4}{5} = \frac{129}{5} = 25.8

Thus, the mean is 25.8.

Answer:

Let us choose a = 40, h = 10, then d_i = x_i − 40 and u_{i = $\frac{x_{i} - 40}{10}$}.

Using Step-deviation method, the given data is shown as follows:

Class	Frequency *(f_i)*	*Class mark (x_i)*	d_i = x_i − 40	u_{i = $\frac{x_{i} - 40}{10}$}	*f_iu_i*
5−15	6	10	−30	−3	−18
15−25	10	20	−20	−2	−20
25−35	16	30	−10	−1	−16
35−45	15	40	0	0	0
45−55	24	50	10	1	24
55−65	8	60	20	2	16
65−75	7	70	30	3	21
Total	∑ f_i = 86				∑ f_iu_i = 7

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 40 + \frac{7}{86} \times 10 = 40 + \frac{70}{86} = 40 + 0.81 = 40.81

Thus, the mean is 40.81.

Answer:

Let us choose a = 202.5, h = 1, then d_i = x_i − 202.5 and u_i = $\frac{x_{i} - 202.5}{1}$ .

Using Step-deviation method, the given data is shown as follows:

Weight (in grams)	Number of packets *(f_i)*	*Class mark (x_i)*	d_i = x_i − 202.5	u_i = $\frac{x_{i} - 202.5}{1}$	*f_iu_i*
200−201	13	200.5	−2	−2	−26
201−202	27	201.5	−1	−1	−27
202−203	18	202.5	0	0	0
203−204	10	203.5	1	1	10
204−205	1	204.5	2	2	2
205−206	1	205.5	3	3	3
Total	∑ f_i = 70				∑ f_iu_i = −38

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 202.5 + (\frac{- 38}{70}) \times 1 = 202.5 - 0.542 = 201.96

Hence, the mean is 201.96 g.

Answer:

Let us choose a = 45, h = 10, then d_i = x_i − 45 and u_{i = $\frac{x_{i} - 45}{10}$}.

Using Step-deviation method, the given data is shown as follows:

Class	Frequency *(f_i)*	*Class mark (x_i)*	d_i = x_i − 45	u_{i = $\frac{x_{i} - 45}{10}$}	*f_iu_i*
20−30	25	25	−20	−2	−50
30−40	40	35	−10	−1	−40
40−50	42	45	0	0	0
50−60	33	55	10	1	33
60−70	10	65	20	2	20
Total	∑ f_i = 150				∑ f_iu_i = −37

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 45 - \frac{37}{150} \times 10 = 45 - \frac{37}{15} = 45 - 2.466 = 42.534

Thus, the mean is 42.534.

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 33)}{6}$	$(f_{i} \times u_{i})$
18-24	6	21	−2	−12
24-30	8	27	−1	−8
30-36	12	33 = A	0	0
36-42	8	39	1	8
42-48	4	45	2	8
48-54	2	51	3	6
	$\sum f_{i} = 40$			$\sum (f_{i} \times u_{i}) = 2$

$Now, A = 33, h = 6, \sum f_{i} = 40 and \sum (f_{_{i}} \times u_{i}) = 2 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =33+ {6 \times \frac{2}{40}} =33+0 .3 =33 .3 ∴ x = 33.3 years$

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 550)}{20}$	$(f_{i} \times u_{i})$
500-520	14	510	−2	−28
520-540	9	530	−1	−9
540-560	5	550 = A	0	0
560-580	4	570	1	4
580-600	3	590	2	6
600-620	5	610	3	15
	$\sum f_{i} = 40$			$\sum (f_{i} \times u_{i}) = - 12$

$Now, A = 550, h = 20, \sum f_{i} = 40 and \sum (f_{_{i}} \times u_{i}) = - 12 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =550+ {20 \times \frac{(- 12)}{40}} =550-6 =544 ∴ \bar{x} = 544$

Answer:

Converting the series into exclusive form, we get:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 42)}{5}$	$(f_{i} \times u_{i})$
24.5-29.5	4	27	−3	−12
29.5-34.5	14	32	−2	−28
34.5-39.5	22	37	−1	−22
39.5-44.5	16	42 = A	0	0
44.5-49.5	6	47	1	6
49.5-54.5	5	52	2	10
54.5-59.5	3	57	3	9
	$\sum f_{i} = 70$			$\sum (f_{i} \times u_{i}) = - 37$

$Now, A = 42, h = 5, \sum f_{i} = 70 and \sum (f_{_{i}} \times u_{i}) = - 37 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =42+ {5 \times \frac{(- 37)}{70}} =42-2 .64 =39 .36 ∴ \bar{x} = 39.36 ∴ Mean age=39 .36 years$

Answer:

Converting the series into exclusive form, we get:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 29.5)}{10}$	$(f_{i} \times u_{i})$
4.5-14.5	6	9.5	−2	−12
14.5-24.5	11	19.5	−1	−11
24.5-34.5	21	29.5 = A	0	0
34.5-44.5	23	39.5	1	23
44.5-54.5	14	49.5	2	28
54.5-64.5	5	59.5	3	15
	$\sum f_{i} = 80$			$\sum (f_{i} \times u_{i}) = 43$

$Now, A = 29.5, h = 10, \sum f_{i} = 80 and \sum (f_{_{i}} \times u_{i}) = 43 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =29 .5+ {10 \times \frac{43}{80}} =29 .5+5 .375 =34 .875 ∴ \bar{x} = 34 .875 ∴ The average age of the patients is 34 .87 years .$

Answer:

Let us choose a = 92, h = 5, then d_i = x_i − 92 and u_i = $\frac{x_{i} - 92}{5}$ .

Using Step-deviation method, the given data is shown as follows:

Weight (in grams)	Number of eggs *(f_i)*	*Class mark (x_i)*	d_i = x_i − 92	u_i = $\frac{x_{i} - 92}{5}$	*f_iu_i*
74.5−79.5	4	77	−15	−3	−12
79.5−84.5	9	82	−10	−2	−18
84.5−89.5	13	87	−5	−1	−13
89.5−94.5	17	92	0	0	0
94.5−99.5	12	97	5	1	12
99.5−104.5	3	102	10	2	6
104.5−109.5	2	107	15	3	6
Total	∑ f_i = 60				∑ f_iu_i = −19

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 92 + (\frac{- 19}{60}) \times 5 = 92 - 1.58 = 90.42 \approx 90

Thus, the mean weight to the nearest gram is 90 g.

Answer:

Let us choose a = 17.5, h = 5, then d_i = x_i − 17.5 and u_i = $\frac{x_{i} - 17.5}{5}$ .

Using Step-deviation method, the given data is shown as follows:

Marks	Number of students (cf)	Frequency (f_i)	*Class mark (x_i)*	d_i = x_i − 17.5	u_i = $\frac{x_{i} - 17.5}{5}$	*f_iu_i*
0−5	3	3	2.5	−15	−3	−9
5−10	10	7	7.5	−10	−2	−14
10−15	25	15	12.5	−5	−1	−15
15−20	49	24	17.5	0	0	0
20−25	65	16	22.5	5	1	16
25−30	73	8	27.5	10	2	16
30−35	78	5	32.5	15	3	15
35−40	80	2	37.5	20	4	8
Total		∑ f_i = 80				∑ f_iu_i = 17

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 17.5 + (\frac{17}{80}) \times 5 = 17.5 + 1.06 = 18.56

Thus, the mean marks correct to 2 decimal places is 18.56.

Answer:

We prepare the cumulative frequency table, as shown below:

Age (in years)	Number of patients (*f_i*)	Cumulative Frequency (cf)
0−15	5	5
15−30	20	25
30−45	40	65
45−60	50	115
60−75	25	140
Total	N = ∑ f_i = 140

Now, N = 140

\Rightarrow \frac{N}{2} = 70

.

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.

Thus, the median class is 45−60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 45 + (\frac{\frac{140}{2} - 65}{50}) \times 15 = 45 + (\frac{70 - 65}{50}) \times 15 = 45 + 1.5 = 46.5

Hence, the median age is 46.5 years.

Answer:

Class	Frequency (f)	Cumulative frequency
0-7	3	3
7-14	4	7
14-21	7	14
21-28	11	25
28-35	0	25
35-42	16	41
42-49	9	50
	N=∑f=50

$\begin{array}{l} N = 50 \\ \Rightarrow \frac{N}{2} = 25 \\ The cumulative frequency just greater than 25 is 41 and the corresponding class is 35-42. \\ Thus, the median class is 35-42. \\ ∴ l = 35, h = 7, f = 16, c f = c .f . of preceding class = 25 and \frac{N}{2} = 25 \end{array} ∴ Median = l + \frac{\frac{N}{2} - c . f}{f} \times h \begin{array}{l} =35+ \{7 \times \frac{(25 - 25)}{16}\} \\ =35+0 \\ =35 \end{array}$

Answer:

Class	Frequency(f)	Cumulative frequency
0-100	40	40
100-200	32	72
200-300	48	120
300-400	22	142
400-500	8	150
	N=∑f=150

$N = 150 \Rightarrow \frac{N}{2} = 75 The cumulative frequency just greater than 75 is 120 and the corresponding class is 200-300. Thus, the median class is 200-300. ∴ l = 200, h = 100, f = 48, c f = c .f . of preceding class=72 and \frac{N}{2} = 75 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} =200+ \{100 \times \frac{(75 - 72)}{48}\} =200+6 .25 =206 .25 Hence, the median daily wage income of the workers is Rs 206 .25.$

Answer:

Class	Frequency(f)	Cumulative frequency
5-10	5	5
10-15	6	11
15-20	15	26
20-25	10	36
25-30	5	41
30-35	4	45
35-40	2	47
40-45	2	49
	N=∑f=49

$N = 49 \Rightarrow \frac{N}{2} = 24.5 The cumulative frequency just greater than 24 .5 is 26 and the corresponding class is 15-20. Thus, the median class is 15-20. Now, l = 15, h = 5, f = 15, c f = c .f . of preceding class = 11 and \frac{N}{2} = 24.5 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} =15+ \{5 \times \frac{(24.5 - 11)}{15}\} =15+4 .5 =19 .5 Hence, median=19 .5$

Answer:

Class	Frequency(f)	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	7	63
185-205	4	67
	N=∑f=67

$N = 67 \Rightarrow \frac{N}{2} = 33.5 The cumulative frequency just greater than 33 .5 is 42 and the corresponding class is 125-145. Thus, the median class is 125-145. Now, l = 125, h = 20, f = 20, c f = c .f . of preceding class = 22 and \frac{N}{2} = 33.5 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} =125+ \{20 \times \frac{(33.5 - 22)}{20}\} =125+11 .5 =136 .5 Hence, median=136 .5$

Answer:

Class	Frequency(f)	Cumulative frequency
135-140	6	6
140=145	10	16
145-150	18	34
150-155	22	56
155-160	20	76
160-165	15	91
165-170	6	97
170-175	3	100
	N=∑f=100

$N = 100 \Rightarrow \frac{N}{2} = 50 The cumulative frequency just greater than 50 is 56 and the corresponding class is 150-155. Thus, the median class is 150-155. Now, l = 150, h = 5, f = 22, c f = c .f . of preceding class=34 and \frac{N}{2} = 50 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} =150+ \{5 \times \frac{(50 - 34)}{22}\} =150+3 .64 =153 .64 Hence, median=153 .64$

Answer:

Class	Frequency (f_i)	c.f
0-10	5	5
10-20	25	30
20-30	x	x+30
30-40	18	x+48
40-50	7	x+55

Median is 24 which lies in 20 - 30 ∴ Median Class = 20 - 30 Let the unknown frequency be x Here, l = 20, \frac{n}{2} = \frac{x + 55}{2}, c . f of the preceding class = c . f = 30, f = x, h = 10 ∴ Median = l + \frac{\frac{n}{2} - c . f}{f} \times h \Rightarrow 24 = 20 + \frac{\frac{x + 55}{2} - 30}{x} \times 10 \Rightarrow 24 = 20 + \frac{\frac{x + 55 - 60}{2}}{x} \times 10 \Rightarrow 24 = 20 + \frac{x - 5}{2 x} \times 10 \Rightarrow 24 = 20 + \frac{5 x - 25}{x} \Rightarrow 24 = \frac{20 x + 5 x - 25}{x} \Rightarrow 24 x = 25 x - 25 \Rightarrow - x = - 25 \Rightarrow x = 25 Hence, the unknown frequency is 25

Answer:

We prepare the cumulative frequency table, as shown below:

Class	*Frequency (f_i)*	Cumulative frequency (cf)
0−5	12	12
5−10	a	12 + a
10−15	12	24 + a
15−20	15	39 + a
20−25	b	39 + a + b
25−30	6	45 + a + b
30−35	6	51 + a + b
35−40	4	55 + a + b
Total	N = ∑f_i = 70

Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,

55 + a + b = 70 ⇒ a + b = 15 ...(1)

Median is 16, which lies in 15−20. So, the median class is 15−20.

∴ l = 15, h = 5, N = 70, f = 15 and cf = 24 + a

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h \Rightarrow 16 = 15 + (\frac{\frac{70}{2} - (24 + a)}{15}) \times 5 \Rightarrow 16 = 15 + (\frac{35 - 24 - a}{3}) \Rightarrow 16 = 15 + (\frac{11 - a}{3}) \Rightarrow 16 - 15 = \frac{11 - a}{3} \Rightarrow 1 \times 3 = 11 - a \Rightarrow a = 11 - 3 \Rightarrow a = 8

∴ b = 15 − a [From (1)]
⇒ b = 15 − 8
⇒ b = 7

Hence, a = 8 and b = 7.

Answer:

We prepare the cumulative frequency table, as shown below:

Runs scored	Number of batsmen *(f_i)*	Cumulative frequency (cf)
2500−3500	5	5
3500−4500	x	5 + x
4500−5500	y	5 + x + y
5500−6500	12	17 + x + y
6500−7500	6	23 + x + y
7500−8500	2	25 + x + y
Total	N = ∑f_i = 60

Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,

25 + x + y = 60 ⇒ x + y = 35 ...(1)

Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h \Rightarrow 5000 = 4500 + (\frac{\frac{60}{2} - (5 + x)}{y}) \times 1000 \Rightarrow 5000 - 4500 = (\frac{30 - 5 - x}{y}) \times 1000 \Rightarrow 500 = (\frac{25 - x}{y}) \times 1000 \Rightarrow y = 50 - 2 x \Rightarrow 35 - x = 50 - 2 x [From (1)] \Rightarrow 2 x - x = 50 - 35 \Rightarrow x = 15

∴ y = 35 − x [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.

Answer:

Class	Frequency(f)	Cumulative frequency
0-10	$f_{1}$	$f_{1}$
10-20	5	$f_{1}$ +5
20-30	9	$f_{1}$ +14
30-40	12	$f_{1}$ +26
40-50	$f_{2}$	$f_{1} + f_{2}$ +26
50-60	3	$f_{1} + f_{2}$ +29
60-70	2	$f_{1} + f_{2}$ +31
	N=∑f=40

$\begin{array}{l} Now, f_{1} + f_{2} + 31 = 40 \\ \Rightarrow f_{1} + f_{2} = 9 \\ \Rightarrow f_{2} = 9 - f_{1} .. . (i) \\ The median is 32 .5 which lies in 30-40. \\ Hence, median class = 30 - 40 \\ Here, l = 30, \frac{N}{2} = \frac{40}{2} = 20, f = 12 and c . f = 14 + f_{1} \\ Now, median = 32.5 \\ \Rightarrow l + \frac{\frac{N}{2} - c . f}{f} \times h = 32.5 \\ \Rightarrow 30 + \frac{20 - (14 + f_{1})}{12} \times 10 = 32.5 \end{array} \Rightarrow \frac{6 - f_{1}}{12} \times 10 = 2.5 \begin{array}{l} \Rightarrow \frac{60 - 10 f_{1}}{12} = 2.5 \\ \Rightarrow 60 - 10 f_{1} = 30 \\ \Rightarrow 10 f_{1} = 30 \\ \Rightarrow f_{1} = 3 \\ From equation (i), we have: \\ f_{2} = 9 - 3 \\ \Rightarrow f_{2} = 6 \end{array}$

Answer:

First, we will convert the data into exclusive form.

Class	Frequency(f)	Cumulative frequency
18.5-25.5	35	35
25.5-32.5	96	131
32.5-39.5	68	199
39.5-46.5	102	301
46.5-53.5	35	336
53.5-60.5	4	340
	N=∑f=340

$N = 340 = > \frac{N}{2} = 170 The cumulative frequency just greater than 170 is 199 and the corresponding class is 32 .5-39 .5. Thus, the median class is 32 .5-39 .5. ∴ l = 32.5, h = 7, f = 68, c f = c .f . of preceding class = 131 and \frac{N}{2} = 170 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} = 32 .5+ \{7 \times \frac{(170 - 131)}{68}\} = 32 .5+4 .01 = 36 .51 Hence, median = 36 .51$

Answer:

Converting the given data into exclusive form, we get:

Class	Frequency(f)	Cumulative frequency
60.5-70.5	5	5
70.5-80.5	15	20
80.5-90.5	20	40
90.5-100.5	30	70
100.5-110.5	20	90
110.5-120.5	8	98
	N=∑f=98

$N = 98 \Rightarrow \frac{N}{2} = 49 The cumulative frequency just greater than 49 is 70 and the corresponding class is 90 .5-100 .5. Thus, the median class is 90 .5-100 .5. Now, l = 90.5, h = 10, f = 30, c f = c .f . of preceding class = 40 and \frac{N}{2} = 49 ∴ Median, M = l + {h \times \frac{(\frac{N}{2} - c f)}{f}} =90 .5+ {10 \times \frac{(49 - 40)}{30}} =90 .5+3 =93 .5 Hence, median wages = Rs 93 .50$

Answer:

Converting into exclusive form, we get:

Class	Frequency(f)	Cumulative frequency
0.5-5.5	7	7
5.5-10.5	10	17
10.5-15.5	16	33
15.5-20.5	32	65
20.5-25.5	24	89
25.5-30.5	16	105
30.5-35.5	11	116
35.5-40.5	5	121
40.5-45.5	2	123
	N=∑f=123

$N = 123 \Rightarrow \frac{N}{2} = 61.5 The cumulative frequency just greater than 61 .5 is 65 and the corresponding class is 15 .5-20 .5. Thus, the median class is 15 .5-20 .5. ∴ l = 15.5, h = 5, f = 32, c f = c .f . of preceding class = 33 and \frac{N}{2} = 61.5 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} =15 .5+ \{5 \times \frac{(61.5 - 33)}{32}\} =15 .5+4 .45 =19 .95 Hence, median=19 .95$

Answer:

Class	Cumulative frequency	Frequency (f)
0-10	12	12
10-20	32	20
20-30	57	25
30-40	80	23
40-50	92	12
50-60	116	24
60-70	164	48
70-80	200	36
		N=∑f=200

$\begin{array}{l} N = 200 \\ \Rightarrow \frac{N}{2} = 100 \\ The cumulative frequency just greater than 100 is 116 and the corresponding class is 50-60 . \\ Thus, the median class is 50-60. \\ ∴ l = 50, h = 10, f = 24, c f = c .f . of preceding class = 92 and \frac{N}{2} = 100 \\ ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} \\ = 50 + \{10 \times \frac{(100 - 92)}{24}\} \\ = 50 + 3.33 \\ = 53.33 \\ Hence, median = 53.33 \end{array}$

Answer:

Here the maximum class frequency is 45, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now,

modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f₁) of the modal class = 45,

frequency (f₀) of class preceding the modal class = 35,

frequency (f₂) of class succeeding the modal class = 25

Now, let us substitute these values in the formula:

$Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 30 + (\frac{45 - 35}{90 - 35 - 25}) \times 10 = 30 + (\frac{10}{30}) \times 10 = 30 + 3.33 = 33.33$

Hence, the mode is 33.33.

Answer:

Here the maximum class frequency is 28, and the class corresponding to this frequency is 40−60. So, the modal class is 40−60.

Now,

Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,

frequency (f₁) of the modal class = 28,

frequency (f₀) of class preceding the modal class = 16,

frequency (f₂) of class succeeding the modal class = 20.

Now, let us substitute these values in the formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 40 + (\frac{28 - 16}{56 - 16 - 20}) \times 20 = 40 + (\frac{12}{20}) \times 20 = 40 + 12 = 52

Hence, the mode is 52.

Answer:

Here the maximum class frequency is 20, and the class corresponding to this frequency is 160−165. So, the modal class is 160−165.

Now,

Modal class = 160−165, lower limit (l) of modal class = 160, class size (h) = 5,

frequency (f₁) of the modal class = 20,

frequency (f₀) of class preceding the modal class = 8,

frequency (f₂) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 160 + (\frac{20 - 8}{40 - 8 - 12}) \times 5 = 160 + (\frac{12}{20}) \times 5 = 160 + 3 = 163

Hence, the mode is 163.

It represents that the height of maximum number of students is 163 cm.

Now, to find the mean let us put the data in the table given below:

Height (in cm)	*Number of students (f_i)*	*Class mark (x_i)*	f_ix_i
150−155	15	152.5	2287.5
155−160	8	157.5	1260
160−165	20	162.5	3250
165−170	12	167.5	2010
170−175	5	172.5	862.5
Total	∑*f_i* = 60		∑*f_i*x_i = 9670

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{9670}{60} = 161.17

Thus, mean of the given data is 161.17.

It represents that on an average, the height of a student is 161.17 cm.

Answer:

As the class 26-30 has the maximum frequency, it is the modal class.

$Now, x_{k} = 26, h = 4, f_{k} = 25, f_{k - 1} = 20, f_{k + 1} = 22 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 26 + \{4 \times \frac{(25 - 20)}{(2 \times 25 - 20 - 22)}\} = 26 + \{4 \times \frac{5}{8}\} = (26 + 2.5) = 28.5$

Answer:

As the class 1500-2000 has the maximum frequency, it is the modal class.
$Now, x_{k} = 1500, h = 500, f_{k} = 40, f_{k - 1} = 24 and f_{k + 1} = 31 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 1500 + \{500 \times \frac{(40 - 24)}{(2 \times 40 - 24 - 31)}\} = 1500 + \{500 \times \frac{16}{25}\} = (1500 + 320) = 1820$

Hence, mode = Rs 1820

Answer:

As the class 5000-10000 has the maximum frequency, it is the modal class.

$Now, x_{k} = 5000, h = 5000, f_{k} = 150, f_{k - 1} = 90 and f_{k + 1} = 100 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 5000 + \{5000 \times \frac{(150 - 90)}{(2 \times 150 - 90 - 100)}\} = 5000 + \{5000 \times \frac{60}{110}\} = (5000 + 2727.27) = 7727.27$

Hence, mode = Rs 7727.27

Answer:

As the class 15-20 has the maximum frequency, it is the modal class.
$Now, x_{k} = 15, h = 5, f_{k} = 24, f_{k - 1} = 18 and f_{k + 1} = 17 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 15 + \{5 \times \frac{(24 - 18)}{(2 \times 24 - 18 - 17)}\} = 15 + \{5 \times \frac{6}{13}\} = (15 + 2.3) = 17.3$

Hence, mode=17.3 years

Answer:

As the class 85-95 has the maximum frequency, it is the modal class.

$Now, x_{k} = 85, h = 10, f_{k} = 32, f_{k - 1} = 30 and f_{k + 1} = 6 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 85 + \{10 \times \frac{(32 - 30)}{(2 \times 32 - 30 - 6)}\} = 85 + \{10 \times \frac{2}{28}\} = (85 + . 71) = 85.71$

Hence, mode=85.71

Answer:

Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below:

Class interval	0.5-5.5	5.5-10.5	10.5-15.5	15.5-20.5	20.5-25.5	25.5-30.5	30.5-35.5	35.5-40.5	40.5-45.5	45.5-50.5
Frequency	3	8	13	18	28	20	13	8	6	4

As the class 20.5-25.5 has the maximum frequency, it is the modal class.

Now, x_{k} = 20.5, h = 5, f_{k} = 28, f_{k - 1} = 18 and f_{k + 1} = 20 ∴ Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 20.5 + \{5 \times \frac{(28 - 18)}{(2 \times 28 - 18 - 20)}\} = 20.5 + \{5 \times \frac{10}{18}\} = (20.5 + 2.78) = 23.28

Hence, mode=23.28

Answer:

It is given that the sum of frequencies is 181.

∴ x + 15 + 18 + 30 + 50 + 48 + x = 181
⇒ 2x + 161 = 181
⇒ 2x = 181 − 161
⇒ 2x = 20
⇒ x = 10

Thus, x = 10.

Here the maximum class frequency is 50, and the class corresponding to this frequency is 13−15. So, the modal class is 13−15.

Now,

Modal class = 13−15, lower limit (l) of modal class = 13, class size (h) = 2,

frequency (f₁) of the modal class = 50,

frequency (f₀) of class preceding the modal class = 30,

frequency (f₂) of class succeeding the modal class = 48.

Now, let us substitute these values in the formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 13 + (\frac{50 - 30}{100 - 30 - 48}) \times 2 = 13 + (\frac{20}{22}) \times 2 = 13 + 1.82 = 14.82

Hence, the mode is 14.82.

Answer:

To find the mean let us put the data in the table given below:

Class	Frequency *(f_i)*	*Class mark (x_i)*	f_ix_i
0−10	4	5	20
10−20	4	15	60
20−30	7	25	175
30−40	10	35	350
40−50	12	45	540
50−60	8	55	440
60−70	5	65	325
Total	∑*f_i* = 50		∑*f_i*x_i = 1910

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{1910}{50} = 38.2

Thus, mean of the given data is 38.2.

Now, to find the median let us put the data in the table given below:

Class	Frequency *(f_i)*	Cumulative frequency (cf)
0−10	4	4
10−20	4	8
20−30	7	15
30−40	10	25
40−50	12	37
50−60	8	45
60−70	5	50
Total	N = ∑*f_i* = 50

Now, N = 50

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

.

The cumulative frequency just greater than 25 is 37, and the corresponding class is 40−50.

Thus, the median class is 40−50.

∴ l = 40, h = 10, N = 50, f = 12 and cf = 25.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 40 + (\frac{25 - 25}{12}) \times 10 = 40

Thus, the median is 40.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 40 − 2 × 38.2
          = 120 − 76.4
          = 43.6

Hence, Mean = 38.2, Median = 40 and Mode = 43.6

Answer:

To find the mean let us put the data in the table given below:

Class	Frequency *(f_i)*	*Class mark (x_i)*	f_ix_i
0−20	6	10	60
20−40	8	30	240
40−60	10	50	500
60−80	12	70	840
80−100	6	90	540
100−120	5	110	550
120−140	3	130	390
Total	∑*f_i* = 50		∑*f_i*x_i = 3120

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{3120}{50} = 62.4

Thus, mean of the given data is 62.4.

Now, to find the median let us put the data in the table given below:

Class	Frequency *(f_i)*	Cumulative frequency (cf)
0−20	6	6
20−40	8	14
40−60	10	24
60−80	12	36
80−100	6	42
100−120	5	47
120−140	3	50
Total	N = ∑*f_i* = 50

Now, N = 50

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

.

The cumulative frequency just greater than 25 is 36, and the corresponding class is 60−80.

Thus, the median class is 60−80.

∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 60 + (\frac{25 - 24}{12}) \times 20 = 60 + 1.67 = 61.67

Thus, the median is 61.67.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 61.67 − 2 × 62.4
          = 185.01 − 124.8
          = 60.21

Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21.

Answer:

To find the mean let us put the data in the table given below:

Class	Frequency *(f_i)*	*Class mark (x_i)*	f_ix_i
0−50	2	25	50
50−100	3	75	225
100−150	5	125	625
150−200	6	175	1050
200−250	5	225	1125
250−300	3	275	825
300−350	1	325	325
Total	∑*f_i* = 25		∑*f_i*x_i = 4225

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{4225}{25} = 169

Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:

Class	Frequency *(f_i)*	Cumulative frequency (cf)
0−50	2	2
50−100	3	5
100−150	5	10
150−200	6	16
200−250	5	21
250−300	3	24
300−350	1	25
Total	N = ∑*f_i* = 25

Now, N = 25

\Rightarrow \frac{N}{2} = 12.5

.

The cumulative frequency just greater than 12.5 is 16, and the corresponding class is 150−200.

Thus, the median class is 150−200.

∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 150 + (\frac{12.5 - 10}{6}) \times 50 = 150 + 20.83 = 170.83

Thus, the median is 170.83.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 170.83 − 2 × 169
          = 512.49 − 338
          = 174.49

Hence, Mean = 169, Median = 170.83 and Mode = 174.49

Answer:

To find the mean let us put the data in the table given below:

Marks obtained	Number of students *(f_i)*	*Class mark (x_i)*	f_ix_i
25−35	7	30	210
35−45	31	40	1240
45−55	33	50	1650
55−65	17	60	1020
65−75	11	70	770
75−85	1	80	80
Total	∑*f_i* = 100		∑*f_i*x_i = 4970

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{4970}{100} = 49.7

Thus, mean of the given data is 49.7.

Now, to find the median let us put the data in the table given below:

Class	Frequency *(f_i)*	Cumulative frequency (cf)
25−35	7	7
35−45	31	38
45−55	33	71
55−65	17	88
65−75	11	99
75−85	1	100
Total	N = ∑*f_i* = 100

Now, N = 100

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

.

The cumulative frequency just greater than 50 is 71, and the corresponding class is 45−55.

Thus, the median class is 45−55.

∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 45 + (\frac{50 - 38}{33}) \times 10 = 45 + 3.64 = 48.64

Thus, the median is 48.64.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 48.64 − 2 × 49.70
          = 145.92 − 99.4
          = 46.52

Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52.

Answer:

We have the following:

Height in cm	Mid value $(x_{i})$	Frequency $(f_{i})$	Cumulative frequency	$f_{i} \times x_{i}$
120-130	125	2	2	250
130-140	135	8	10	1080
140-150	145	12	22	1740
150-160	155	20	42	3100
160-170	165	8	50	1320
		∑ $f_{i} = 50$		∑ $(f_{i} \times x_{i}) =$ 7490

Mean,

\bar{x}

=

\frac{\sum (f_{i} \times x_{i})}{\sum f_{i}}

=

\frac{7490}{50}

=149.8

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

The cumulative frequency just greater than 25 is 42 and the corresponding class is 150-160.
Thus, the median class is 150-160.

Now, l = 150, h = 10, f = 20, c = c f

of preceding class = 22 and

\frac{N}{2} = 25

Median, M_{e} = l + \{h \times \frac{(\frac{N}{2} - c)}{f}\} = 150 + \{10 \times \frac{(25 - 22)}{20}\} = (150 + 10 \times \frac{3}{20})

=151.5
∴ Mode = 3(Median)

- 2

(Mean)
= 3

\times 151.5 - 2 \times 149.8

= 154.9

Answer:

We have the following:

Daily income	Mid value $(x_{i})$	Frequency $(f_{i})$	Cumulative frequency	$f_{i} \times x_{i}$
100-120	110	12	12	1320
120-140	130	14	26	1820
140-160	150	8	34	1200
160-180	170	6	40	1020
180-200	190	10	50	1900
		∑ $f_{i} = 50$		∑ $(f_{i} \times x_{i}) =$ 7260

Mean,

\bar{x}

=

\frac{\sum (f_{i} \times x_{i})}{\sum f_{i}}

=

\frac{7260}{50}

=145.2

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

The cumulative frequency just greater than 25 is 26 and the corresponding class is 120-140.
Thus, the median class is 120-140.

N o w, l = 120, h = 20, f = 14, c = c f

of preceding class = 12 and

\frac{N}{2} = 25

Median,

M_{e} = l + \{h \times \frac{(\frac{N}{2} - c)}{f}\}

= 120 + \{20 \times \frac{(25 - 12)}{14}\} = (120 + 20 \times \frac{13}{14})

=138.57
Mode = 3(Median)

-

2(mean)

= 3 \times 138.57 - 2 \times 145.2 = 125. 31

Answer:

We have the following:

Daily income	Mid value	Frequency $(f_{i})$	Cumulative frequency	$f_{i} \times x_{i}$
100-150	125	6	6	750
150-200	175	7	13	1225
200-250	225	12	25	2700
250-300	275	3	28	825
300-350	325	2	30	650
		∑ $f_{i} = 30$		∑ $(f_{i} \times x_{i}) =$ 6150

Mean,

\bar{x}

=

\frac{\sum (f_{i} \times x_{i})}{\sum f_{i}}

=

\frac{6150}{30}

=205

Now, N = 30 \Rightarrow \frac{N}{2} = 15

The cumulative frequency just greater than 15 is 25 and the corresponding class is 200-250.
Thus, the median class is 200-250.

Now, l = 200, h = 50, f = 12, c = c f

of preceding class = 13 and

\frac{N}{2} = 15

Median,

M_{e} = l + \{h \times \frac{(\frac{N}{2} - c)}{f}\}

= 200 + \{50 \times \frac{(15 - 13)}{12}\} = (200 + 50 \times \frac{2}{12})

=200 + 8.33
=208.33

Answer:

The frequency distribution table of less than type is given as follows:

Marks (upper class limits)	Cumulative frequency (cf)
Less than 10	5
Less than 20	5 + 3 = 8
Less than 30	8 + 4 = 12
Less than 40	12 + 3 = 15
Less than 50	15 + 3 = 18
Less than 60	18 + 4 = 22
Less than 70	22 + 7 = 29
Less than 80	29 + 9 = 38
Less than 90	38 + 7 = 45
Less than 100	45 + 8 = 53

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

Here, N = 53

\Rightarrow \frac{N}{2} = 26.5

.

Mark the point A whose ordinate is 26.5 and its x−coordinate is 66.4.

Thus, median of the data is 66.4.

Answer:

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

Here, N = 80 $\Rightarrow \frac{N}{2} = 40$ .

Mark the point A whose ordinate is 40 and its x−coordinate is 76.

Thus, median of the data is 76.

Answer:

The frequency distribution table of more than type is as follows:

Marks (lower class limits)	Cumulative frequency (cf)
More than 0	96 + 4 = 100
More than 10	90 + 6 = 96
More than 20	80 + 10 = 90
More than 30	70 + 10 = 80
More than 40	45 + 25 = 70
More than 50	23 + 22 = 45
More than 60	18 + 5 = 23
More than 70	5

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Answer:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits)	Cumulative frequency (cf)
More than 135	5 + 45 = 50
More than 140	8 + 37 = 45
More than 145	9 + 28 = 37
More than 150	12 + 16 = 28
More than 155	14 + 2 = 16
More than 160	2

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Answer:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits)	Cumulative frequency (cf)
More than 140	3 + 153 = 156
More than 160	8 + 145 = 153
More than 180	15 + 130 = 145
More than 200	40 + 90 = 130
More than 220	50 + 40 = 90
More than 240	30 + 10 = 40
More than 260	10

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Answer:

The frequency distribution table of more than type is as follows:

Production yield (kg/ha) (lower class limits)	Cumulative frequency (cf)
More than 50	2 + 98 = 100
More than 55	8 + 90 = 98
More than 60	12 + 78 = 90
More than 65	24 + 54 = 78
More than 70	38 + 16 = 54
More than 75	16

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Here, N = 100

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

Mark the point A whose ordinate is 50 and its x−coordinate is 70.5.

Thus, median of the data is 70.5.

Answer:

The frequency distribution table of less than type is as follows:

Weekly expenditure (in ₹) (upper class limits)	Cumulative frequency (cf)
Less than 200	5
Less than 300	5 + 6 = 11
Less than 400	11 + 11 = 22
Less than 500	22 + 13 = 35
Less than 600	35 + 5 = 40
Less than 700	40 + 4 = 44
Less than 800	44 + 3 = 47
Less than 900	47 + 2 = 49

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Now,

The frequency distribution table of more than type is as follows:

Weekly expenditure (in ₹) (lower class limits)	Cumulative frequency (cf)
More than 100	44 + 5 = 49
More than 200	38 + 6 = 44
More than 300	27 + 11 = 38
More than 400	14 + 13 = 27
More than 500	9 + 5 = 14
More than 600	5 + 4 = 9
More than 700	2 + 3 = 5
More than 800	2

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Answer:

From the given table, we may prepare the 'more than' frequency table as shown below:

Score	Number of candidates
More than 750	34
More than 700	52
More than 650	79
More than 600	103
More than 550	135
More than 500	175
More than 450	210
More than 400	230

We plot the points A(750,34), B(700,52), C(650,79), D(600,103), E(550,135), F(500,175), G(450,210) and H(400,230).
Join AB, BC, CD, DE, EF, FG, GH and HA with a free hand to get the curve representing the ‘more than type’ series.

Here, N=230
⇒ $\frac{N}{2} = 115$
From P(0,115), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units

Answer:

(i) From the given table, we may prepare the 'less than' frequency table as shown below:

Marks	No. of students
Less than 5	2
Less than 10	7
Less than 15	13
Less than 20	21
Less than 25	31
Less than 30	56
Less than 35	76
Less than 40	94
Less than 45	98
Less than 50	100

We plot the points A(5,2), B(10,7), C(15,13), D(20,21), E(25,31), F(30,56), G(35,76), H(40,94), I(45,98) and J(50,100).
Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.

(ii) More than series:

Marks	No. of student
More than 0	100
More than 5	98
More than 10	93
More than 15	87
More than 20	79
More than 25	69
More than 30	44
More than 35	24
More than 40	6
More than 45	2

Now, on the same graph paper, we plot the points (0,100), (5,98), (10,94), (15,76), (20,56), (25,31), (30,21), (35,13), (40,6) and (45,2).
Join , with a free hand to get the ‘more than type’ series.

The two curves intersect at point L. Draw $LM ⊥ OX$ cutting the $x - a x i s$ at M.
Clearly, M = 29.5
Hence, Median = 29.5

Answer:

(i) Less than series:

Marks	No. of students
Less than 144	3
Less than 148	12
Less than 152	36
Less than 156	67
Less than 160	109
Less than 164	173
Less than 168	248
Less than 172	330
Less than 176	416
Less than 180	450

We plot the points A(144,3), B(148,12), C(152,36), D(156,67), E(160,109) F(164,173), G(168,248), H(172,330), I(176,416) and J(180,450). Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.

(ii) More than series:

Marks	No. of students
More than 140	450
More than 144	447
More than 148	438
More than 152	414
More than 156	383
More than 160	341
More than 164	277
More than 168	202
More than 172	120
More than 176	34

Now on the same graph paper, we plot the points $A_{1}$ (140,450), $B_{1}$ (144,447), $C_{1}$ (148,438), $D_{1}$ (152,414), $E_{1}$ (156,383), $F_{1}$ (160,341), $G_{1}$ (164,277), $H_{1}$ (168,202), $I_{1}$ (172,120) and $J_{1} ($ 176,34).
Join $A_{1} B_{1}, B_{1} C_{1}, C_{1} D_{1}, D_{1} E_{1}, E_{1} F_{1}, F_{1} G_{1}, G_{1} H_{1}, H_{1} I_{1} and I_{1} J_{1}$ with a free hand to get the ‘more than type’ series.

The two curves intersect at point L. Draw $LM ⊥ OX$ cutting the $x - a x i s$ at M.
Clearly, M = 166 cm
Hence, Median = 166 cm

Answer:

To find the median let us put the data in the table given below:

Class	*Frequency (f_i)*	Cumulative frequency (cf)
0−10	4	4
10−20	4	8
20−30	8	16
30−40	10	26
40−50	12	38
50−60	8	46
60−70	4	50
Total	N = ∑f_i = 50

Now, N = 50

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

.

The cumulative frequency just greater than 25 is 26, and the corresponding class is 30−40.

Thus, the median class is 30−40.

Answer:

Here the maximum class frequency is 27, and the class corresponding to this frequency is 40−50. So, the modal class is 40−50.

Now,

Modal class = 40−50, lower limit (l) of modal class = 40

Thus, lower limit (l) of modal class is 40.

Answer:

Here the maximum class frequency is 30, and the class corresponding to this frequency is 150−200. So, the modal class is 150−200.

Also, class mark of the modal class is

\frac{150 + 200}{2} = 175

.

Answer:

If the number of observations is odd, then the median is $(\frac{n + 1}{2}) th$ observation.

Thus, $(\frac{25 + 1}{2}) = 13$ th observation represents the median.

Answer:

There is an empirical relationship between the three measures of central tendency:

3Median = Mode + 2Mean

$\Rightarrow Mean = \frac{3 Median - Mode}{2} = \frac{3 (1250) - 1000}{2} = 1375$

Thus, the mean is 1375.

Answer:

Here the maximum class frequency is 25, and the class corresponding to this frequency is 40−60.

So, the modal class is 40−60.

Now, to find the median class let us put the data in the table given below:

Marks obtained	*Number of students (f_i)*	Cumulative frequency (cf)
0−20	4	4
20−40	6	10
40−60	25	35
60−80	10	45
80−100	5	50
Total	N = ∑*f_i* = 50

Now, N = 50

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

.

The cumulative frequency just greater than 25 is 35, and the corresponding class is 40−60.

Thus, the median class is 40−60.

Answer:

Class mark = $\frac{Upper limit + Lower limit}{2}$

∴ class mark of 10−25 = $\frac{10 + 25}{2}$
= 17.5

and class mark of 35−55 = $\frac{35 + 55}{2}$
= 45

Answer:

According to assumed-mean method,

$\bar{x} = A + \frac{\sum_{i} f_{i} d_{i}}{\sum_{i} f_{i}} = 25 + \frac{110}{50} = 25 + 2.2 = 27.2$

Thus, mean is 27.2.

Answer:

According to the question,

4 = $\frac{X}{36}$ and 3 = $\frac{Y}{64}$

⇒ X = 4 × 36 and Y = 3 × 64

⇒ X = 144 and Y = 192

Now, X + Y = 144 + 192 = 336

and total number of observations = 36 + 64 = 100

Thus, mean = $\frac{336}{100} = 3.36$ .

Answer:

Uppar class boundary = Lowest class boundary + width × number of classes

= 8.1 + 2.5 × 12

= 8.1 + 30

= 38.1

Thus, upper class boundary of the highest class is 38.1.

Answer:

If number of observations is even, then the median will be the average of $(\frac{n}{2})$ th and the $(\frac{n}{2} + 1)$ th observations.

In the given case, n = 10 $\Rightarrow (\frac{n}{2}) th = 5 th and (\frac{n}{2} + 1) th = 6 th$ observation.

Thus, $63 = \frac{x + (x + 2)}{2}$
$\Rightarrow 126 = 2 x + 2 \Rightarrow 124 = 2 x \Rightarrow x = 62$

Thus, the value of x is 62.

Answer:

Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged.

Thus, the median of 21 observations taken together is 30.

Answer:

Arranging the observations in ascending order, we have

$\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x$ .

Thus, the median is $\frac{x}{3}$ .

$\Rightarrow \frac{x}{3} = 8 \Rightarrow x = 3 \times 8 \Rightarrow x = 24$

Thus, the value of x is 24.

Answer:

Here the maximum class frequency is 23, and the class corresponding to this frequency is 12−15.

So, the modal class is 12−15.

Now, to find the cumulative frequency let us put the data in the table given below:

Class	Frequency *(f_i)*	Cumulative frequency (cf)
3−6	7	7
6−9	13	20
9−12	10	30
12−15	23	53
15−18	4	57
18−21	21	78
21−24	16	94
Total	N = ∑*f_i* = 94

Thus, the cumulative frequency of the modal class is 53.

Answer:

Here the maximum class frequency is 18, and the class corresponding to this frequency is 40−60.

So, the modal class is 40−60.

Now,

Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,

frequency (f₁) of the modal class = 18,

frequency (f₀) of class preceding the modal class = 6,

frequency (f₂) of class succeeding the modal class = 10.

Now, let us substitute these values in the formula:

$Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 40 + (\frac{18 - 6}{36 - 6 - 10}) \times 20 = 40 + (\frac{12}{20}) \times 20 = 40 + 12 = 52$

Hence, the mode is 52.

Answer:

A 'less than type' cumulative frequency distribution table is given below:

Age (in years)	Cumulative frequency (cf)
Less than 20	60
Less than 30	102
Less than 40	157
Less than 50	227
Less than 60	280
Less than 70	300

Answer:

Here, p = 11 + 12 = 23

and 33 + q = 46 ⇒ q = 46 − 33 = 13

Thus, p is 23 and q is 13.

Now,

Here the maximum class frequency is 20, and the class corresponding to this frequency is 500−600.

So, the modal class is 500−600.

Also, ∑ f = N = 80 $\Rightarrow \frac{N}{2} = 40 .$

The cumulative frequency just greater than 40 is 46, and the corresponding class is 400−500.

Thus, the median class is 400−500.

Answer:

The cumulative frequency distribution table of more than type is as follows:

Monthly consumption (in units) (lower class limits)	Cumulative frequency (cf)
More than 65	60 + 4 = 64
More than 85	55 + 5 = 60
More than 105	42 + 13 = 55
More than 125	22 + 20 = 42
More than 145	8 + 14 = 22
More than 165	8

Answer:

The frequency distribution is as follows:

Life-time (in days)	Frequency (f)
0−50	7
50−100	14
100−150	31
150−200	27
200−250	12
250−300	9

Answer:

(a) The frequency distribution into the continuous form is as follows:

Marks obtained (in per cent)	Number of students (f)
10.5−20.5	141
20.5−30.5	221
30.5−40.5	439
40.5−50.5	529
50.5−60.5	495
60.5−70.5	322
70.5−80.5	153

(b) Now, to find the median class let us put the data in the table given below:

Marks obtained (in per cent)	Number of students (f)	Cumulative frequency (cf)
10.5−20.5	141	141
20.5−30.5	221	362
30.5−40.5	439	801
40.5−50.5	529	1330
50.5−60.5	495	1825
60.5−70.5	322	2147
70.5−80.5	153	2300

Now, N = 2300

\Rightarrow \frac{N}{2} = 1150

.

The cumulative frequency just greater than 1150 is 1330, and the corresponding class is 40.5−50.5.

Thus, the median class is 40.5−50.5.

Now, class mark =

\frac{upper class limit + lower class limit}{2}

=

\frac{40.5 + 50.5}{2} = \frac{91}{2} = 45.5

Thus, class mark of the median class is 45.5.

(c) Here the maximum class frequency is 529, and the class corresponding to this frequency is 40.5−50.5.

So, the modal class is 40.5−50.5 and its cumulative frequency is 1330.

Answer:

The given data is shown as follows:

Class	Frequency *(f_i)*	*Class mark (x_i)*	*f_i*x_i
0−10	8	5	40
10−20	p	15	15p
20−30	12	25	300
30−40	13	35	455
40−50	10	45	450
Total	∑ *f_i=* 43 + p		∑ *f_i*x_i= 1245 + 15p

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 27 = \frac{1245 + 15 p}{43 + p} \Rightarrow 1161 + 27 p = 1245 + 15 p \Rightarrow 27 p - 15 p = 1245 - 1161 \Rightarrow 12 p = 84 \Rightarrow p = 7

Thus, the value of p is 7.

Answer:

Let the missing frequency be x.

To find the median let us put the data in the table given below:

Age (in years)	Number of persons (f)	Cumulative frequency (cf)
0−10	5	5
10−20	25	30
20−30	x	30 + x
30−40	18	48 + x
40−50	7	55 + x

The given median is 24,

∴ the median class is 20−30.

∴ l = 20, h = 10, N = 55 + x, f = x and cf = 30

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h \Rightarrow 24 = 20 + (\frac{\frac{55 + x}{2} - 30}{x}) \times 10 \Rightarrow 24 - 20 = (\frac{55 + x - 60}{2 x}) \times 10 \Rightarrow 4 = (\frac{x - 5}{2 x}) \times 10 \Rightarrow 8 x = 10 x - 50 \Rightarrow 2 x = 50 \Rightarrow x = 25

Thus, the missing frequency is 25.

Answer:

(d) Standard deviation

The standard deviation is a measure of dispersion. It is the action or process of distributing things over a wide area (nothing about central location).

Answer:

(a) Mean

The mean can not be determined graphically because the values cannot be summed.

Answer:

Mean is influenced by extreme values.

Hence, the correct answer is option (a).

Answer:

The correct option is (c).

The mode of a frequency distribution can be obtained graphically from a histogram.

Answer:

(d) ogives

This is because median of a frequency distribution is found graphically with the help of ogives.

Answer:

The cumulative frequency table is useful in determining the (b) median.

Answer:

The abscissa of the point of intersection of the 'less than type' and that of the 'more than type' cumulative frequency curves of a grouped data gives its (b) median.

Answer:

$\begin{array}{l} We know that \bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \\ \Rightarrow \bar{x} \sum f_{i} = \sum f_{i} x_{i} . . . (i) \\ Now, \sum f_{i} (x_{i} - \bar{x}) = \sum f_{i} x_{i} - \bar{x} \sum f_{i} \\ \Rightarrow \sum f_{i} (x_{i} - \bar{x}) = \sum f_{i} x_{i} - \sum f_{i} x_{i} [U sing (i)] \\ \Rightarrow \sum f_{i} (x_{i} - \bar{x}) = 0 \\ Hence, the correct option is (b) . \end{array}$

Answer:

(b) $u_{i} = \frac{(x_{i} - A)}{h}$

Answer:

The _{$d_{i}$ 's} are the deviations from $A$ of (c) midpoints of the classes.

Answer:

While computing the mean of the group data, we assume that the frequencies are (b) centred at the class marks of the classes.

Answer:

(b) $mode = (3 \times median) - (2 \times mean)$

Answer:

The x−coordinate represents the median of the given data.

Thus, median of the given data is 20.5.

Hence, the correct answer is option (c).

Answer:

(b) 315
The class having the maximum frequency is the modal class.

$So, the modal class is 150 - 155 and its lower limit is 150 . Also, N = 60 \Rightarrow \frac{N}{2} = 30 The cumulative freequency just more than 30 is 37 and its class is 160 - 165, whose upper limit is 165 . ∴ Required sum = (150 + 165) = 315$

Answer:

$(c) 30 - 40 . The class 30 - 40 has the maximum freequency, i . e ., 30 . So, the modal class is 30 - 40 .$

Answer:

$(b) x_{k} + h \{\frac{f_{k} - f_{k - 1}}{2 f_{k} - f_{k - 1} - f_{k + 1}}\}$

Answer:

$(a) l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\}$

Answer:

$(c) 9.2 It is given that the mean and median are 8.9 a n d 9, r e s p e c t i v e l y . ∴ M o d e = (3 \times M e d i a n) - (2 \times M e a n) \Rightarrow M o d e = (3 \times 9) - (2 \times 8.9) = 27 - 17.8 = 9.2$

Answer:

(b) 57.5

$Class interval$	$35 - 45$	$45 - 55$	$55 - 65$	$65 - 75$
$Frequency$	$8$	$12$	$20$	$10$
$Cumulative frequency$	$8$	$20$	$40$	$50$

Here, N = 50 \Rightarrow \frac{N}{2} = 25, w h i c h l i e s i n t h e c l a s s i n t e r v a l o f 55 - 65 . Now, c f = 55, f = 20 and l = 50 ∴ Median = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} = 50 + \frac{65 - 55}{20} \times (25 - 20) = 57.5

Answer:

(c) 24.4
The maximum frequency is 25 and the modal class is 22-26.

$Now, x_{k} = 22, f_{k} = 25, f_{k - 1} = 16, f_{k + 1} = 19 a n d h = 4 ∴ Mode = x_{k} + h \times \{\frac{f_{k} - f_{k - 1}}{2 f_{k} - f_{k - 1} - f_{k +!}}\} = 22 + 4 \times \frac{(25 - 16)}{(2 \times 25 - 16 - 19)} = 22 + 4 \times \frac{(25 - 16)}{(50 - 16 - 19)} = 22 + 4 \times \frac{9}{15} = 22 + \frac{12}{5} = 22 + 2.4 = 24.4$

Answer:

(c) 24
$Mode = (3 \times median) - (2 \times mean) \Rightarrow (3 \times Median) = (mode + 2 mean) \Rightarrow (3 \times Median) = 16 + 56 \Rightarrow (3 \times Median) = 72 \Rightarrow Median = \frac{72}{3} ∴ Median = 24$

Answer:

(b) 24.5
$Mode = (3 \times median) - (2 \times mean) \Rightarrow (2 \times Mean) = (3 \times median) - mode \Rightarrow (2 \times Mean) = 3 \times 26 - 29 \Rightarrow (2 \times Mean) = 49 \Rightarrow Mean = \frac{49}{2} ∴ Mean = 24.5$

Answer:

$(c) mean = mode = median$
A symmetric distribution is one where the left and right hand sides of the distribution are roughly equally balanced around the mean.

Answer:

Converting the given data into a frequency table, we get:

Monthly income	No. of families	Frequency
30,000 and Above	15	15
25,000-30,000	37	(37 − 15) = 22
20,000-25,000	50	(50 − 37) = 13
18,000-20,000	69	(69 − 50) = 19
14,000-18,000	85	(85 − 69) = 16
10,000-14,000	100	(100 − 85) = 15

Hence, the number of families having an income range of Rs 20,000-Rs 25,000 is 13.
The correct option is (c).

Answer:

First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19.

Median of 8 numbers is average of 4th and 5th terms.

i.e. average of 7 and 11

Thus, the median is 9.

Hence, the correct answer is option (b).

Answer:

It is given that mean of 20 numbers is zero.

i.e. average of 20 numbers is zero.

i.e. sum of 20 numbers is zero.

Thus, at most, there can be 19 positive numbers.
(such that if sum of 19 positive numbers is x, 20th number will be −x)

Hence, the correct answer is option (d).

Answer:

Median of 6 numbers is the average of 3rd and 4th term.

$∴ 13 = \frac{(x - 1) + (x - 3)}{2} \Rightarrow 26 = 2 x - 4 \Rightarrow 2 x = 30 \Rightarrow x = 15$

Thus, x is equal to 15.

Hence, the correct answer is option (c).

Answer:

$Mean = \frac{Sum of observations}{Number of observations} \Rightarrow 15 = \frac{2 + 7 + 6 + x}{4} \Rightarrow 60 = 15 + x \Rightarrow x = 45 . . . . (1)$

Now,

$Mean = \frac{Sum of observations}{Number of observations} \Rightarrow 10 = \frac{18 + 1 + 6 + x + y}{5} \Rightarrow 50 = 25 + x + y \Rightarrow y = 25 - x \Rightarrow y = 25 - 45 [From (1)] \Rightarrow y = - 20$

Disclaimer : There is a misprinting in the question. If the mean of 2, 7, 6 and x is 15 then the value of y be −20 and if the mean of 2, 7, 6 and x is 5 then the value of y be 20. So, the correct answer is option (c).

Answer:

Column I	Column II
(a) The most frequent value in a data is known as ........ .	(s) mode
(b) Which of the following cannot be determined graphically out of mean, mode and median?	(r) mean
(c) An ogive is used to determine ....... .	(q) median
(d) Out of mean, mode, median and standard deviation, which is not a measure of central tendency?	(p) standard deviation

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Clearly, reason (R) is true.
Using the relation given in reason (R), we have:
$2 mean = (3 \times median) - mode = (3 \times 150) - 154 = 450 - 154 = 296 \Rightarrow Mean = 148, which is true . ∴ This assertion (A) and reason (R) are both true and reason (R) is the correct explanation of assertion (A) .$

Answer:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is a not a correct explanation of Assertion (A).
Clearly, reason (R) is true.
The maximum frequency is 23 and the modal class is 12-15.
$Now, x_{k} = 12, f_{k} = 23, f_{k - 1} = 21, f_{k + 1} = 23 and h = 3 ∴ Mode = \{12 + 3 \times \frac{(23 - 21)}{(2 \times 23 - 21 - 10)}\} = (12 + 3 \times \frac{2}{15}) = (12 + 0.4) = 12 ∴ A s s e r t i o n (A) i s t r u e . However, r e a s o n (R) i s n o t a c o r r e c t e x p l a n a t i o n o f a s s e r t i o n (A) .$

Answer:

The cumulative frequency table is useful in determining the median.

Hence, the correct option is (b).

Answer:

(c) 45
$Here, mean = 27 and mode = 33 ∴ Mode = 3 Median - 2 Mean = 3 \times 33 - 2 \times 27 = 99 - 54 = 45$

Answer:

(b) 25

Class	Frequency	Cumulative frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	63

Now, N=63
$∴ \frac{N}{2} = 32.5$
The cumulative frequency just greater than 32.5 is 37 and the corresponding class is 10-15.
$∴ Median class = 10 - 15$
Here, the highest frequency is 20 and its corresponding class is 15-20.
$∴ Modal Class = 15 - 20$
Sum of the lower limits of the median class and modal class = 10 + 15 = 25

Answer:

(d) 17.5
Converting the given series into continuous series, we get:

Class	Frequency	Cumulative frequency
0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

Now, N =57
$∴ \frac{N}{2} = 28.5$

The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.
∴ The median class is 11.5-17.5 and the corresponding upper limit is 17.5.

Answer:

Given:
$Mean = 53.4 Mode = 55.2 We know that mode = 3 median - 2 mean \Rightarrow 55.2 = 3 median - 2 \times 53.4 \Rightarrow 3 Median = 55.2 + 106.8 \Rightarrow Median = \frac{162}{3} = 54$

Answer:

Given distribution table can be written as following:

Class	Frequency	Cumulative frequency
Less than 14	2	2
Less than 14.2	4	6
Less than 14.4	15	21
Less than 14.6	54	75
Less than 14.8	25	100
Less than 15	20	120

Number of athletes who completed the race in less than 14.6 sec = 75

Answer:

Class	Frequency	Cumulative frequency
0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

N=57
$∴ \frac{N}{2} = 28.5$
The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.

$Now, we have : Median class = 11.5 - 17.5 Upper limit = 17.5$

Answer:

The frequency distribution table of the given data is as follows:

Profit (in lakhs ₹)	Number of shops
5−10	30 − 28 = 2
10−15	28 − 16 = 12
15−20	16 − 14 = 2
20−25	14 − 10 = 4
25−30	10 − 7 = 3
30−35	7 − 3 = 4
35−40	3

Thus, the frequency corresponding to the class 20−25 is 4.

Answer:

We have the following table:

Class	Mid value $(x_{i})$	Frequency $(f_{i})$	$f_{i} x_{i}$
1-3	2	9	18
3-5	4	22	88
5-7	6	27	162
7-9	8	18	144
		$\sum f_{i} = 76$	$\sum f_{i} x_{i} = 412$

$∴ Mean, \bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{412}{76} = 5.42$

Answer:

Speed (km/h)	No. of players $(f_{i})$	c.f
85-100	10	10
100-115	4	14
115-130	7	21
130-145	9	30

N =30
$∴ \frac{N}{2} = 15$
The cumulative frequency just greater than 15 is 21.
$∴ Median Class = 115 - 130$
$i . e ., x_{k} = 115, h = 15, f = 7, c = c . f of the preceeding class = 14, \frac{N}{2} = 15$
$∴ M_{e} = x_{k} + \{h \times \frac{(\frac{N}{2} - c)}{f}\} = 115 + (15 \times \frac{15 - 14}{7}) = 115 + 2.1 = 117.1$
Hence, the required speed is 117.1 km/h.

Answer:

Class	Mid value $(x_{i})$	Frequency $(f_{i})$	$(x_{i} \times f_{i})$
0-10	5	16	80
10-20	15	$p$	15 $p$
20-30	25	30	750
30-40	35	32	1120
40-50	45	14	630
		$\sum f_{i} = p + 92$	$\sum f_{i} x_{i} = 15 p + 2580$

$∴ Mean, \bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{15 p + 2580}{p + 92} Now, \frac{15 p + 2580}{p + 92} = 50 \Rightarrow 50 p + 4600 = 15 p + 2580 \Rightarrow 35 p = - 2020 \Rightarrow p = - \frac{404}{7} (frequency cannot be negative)$

So, there is an error in question.

Answer:

As the class 20-30 has the maximum frequency, it is the modal class.

$∴ x_{k} = 20, h = 10, f_{k} = 30, f_{k - 1} = 16 and f_{k + 1} = 9 Mode, M_{0} = x_{k} + \{h \times \frac{(f_{k} - f_{k - 1})}{(2 f_{k} - f_{k - 1} - f_{k + 1})}\} = 20 + \{10 \times \frac{(30 - 16)}{(2 \times 30 - 16 - 9)}\} = 20 + \{10 \times \frac{14}{35}\} = (20 + 4) = 24$

Answer:

Let us plot the points A(10,3), B(20,11), C(30,28), D(40,48) and E(50,70).
Now, let us join AB, BC, CD and DE with a free hand to get the curve representing the ‘less than type’ series.

Answer:

To find the median let us put the data in the table given below:

Class interval	*Frequency (f_i)*	Cumulative frequency (cf)
0−10	8	8
10−20	16	24
20−30	36	60
30−40	34	94
40−50	6	100
Total	N = ∑ f_i = 100

Now, N = 100

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

The cumulative frequency just greater than 50 is 60, and the corresponding class is 20−30.

Thus, the median class is 20−30.

∴ l = 20, h = 10, N = 100, f = 36 and cf = 24.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 20 + (\frac{50 - 24}{36}) \times 10 = 20 + \frac{260}{36} = 27.22

Thus, the median is 27.22.

Answer:

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

Here, N = 100

Mark the point A whose ordinate is 50 and its x−coordinate is 55.

Thus, the median of the data is 55.

Answer:

Let, $f_{1} and f_{2}$ be the frequencies of the class intervals 20-30 and 40-50, respectively.
$Then 10 + 20 + f_{1} + 40 + f_{2} + 25 + 15 = 170 \Rightarrow f_{1} + f_{2} = 60$
The median is 35 which lies in the class of 30-40. So, the median class is 30-40.
$Now, l = 30, h = 10, f = 40, N = 170 and c f = 10 + 20 + f_{1} = f_{1} + 30 ∴ Median, M = l + \{h \times \frac{(\frac{N}{2} - c f)}{f}\} \Rightarrow 30 + [10 \times \frac{85 - (f_{1} + 30)}{40}] = 35 \Rightarrow 30 + \frac{55 - f_{1}}{4} = 35 \Rightarrow 55 - f_{1} = 20 \Rightarrow f_{1} = 35 Now, f_{2} = (60 - 35) = 25 Hence, f_{1} = 35 a n d f_{2} = 25$

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-20	17	10	170
20-40	f₁	30	30 f₁
40-60	32	50	1600
60-80	52- f₁	70	3640-70 f₁
80-100	19	90	1710
	$\sum f_{i} = 120$		_{$\sum (f_{i} \times x_{i}) = 7120 - 40 f_{1}$}

$We have: 17+ f_{1} + 32 + f_{2} + 19 = 120 \Rightarrow f_{1} + f_{2} = 52 \Rightarrow f_{2} = 52 - f_{1} ∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} \Rightarrow 50= \frac{7120 - 40 f_{1}}{120} [∵ Mean=50] \Rightarrow 40 f_{1} = 1120 ∴ f_{1} = 28 And f_{2} = 52 - 28 \Rightarrow f_{2} =24 ∴ The missing frequencies are f_{1} = 28 and f_{2} =24.$

Answer:

Let us choose a = 105, h = 6, then d_i = x_i − 105 and u_i= $\frac{x_{i} - 105}{6}$ .

Using step-deviation method, the given data is shown as follows:

Class	*Frequency (f_i)*	*Class mark (x_i)*	d_i = x_i − 105	u_i= $\frac{x_{i} - 105}{6}$	*f_iu_i*
84−90	15	87	−18	−3	−45
90−96	22	93	−12	−2	−44
96−102	20	99	−6	−1	−20
102−108	18	105	0	0	0
108−114	20	111	6	1	20
114−120	25	117	12	2	50
Total	∑ f_i = 120				∑ f_iu_i = −39

The mean of the given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 105 + (\frac{- 39}{120}) \times 6 = 105 - 1.95 = 103.05

Thus, mean of the given data is 103.05.

Answer:

Here the maximum class frequency is 15, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now,

Modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f₁) of the modal class = 15,

frequency (f₀) of class preceding the modal class = 10,

frequency (f₂) of class succeeding the modal class = 5.

Now, let us substitute these values in the formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 30 + (\frac{15 - 10}{30 - 10 - 5}) \times 10 = 30 + (\frac{50}{15}) = 33.33

Hence, the mode is 33.33.

Now, to find the mean let us put the data in the table given below:

Class	*Frequency (f_i)*	*Class mark (x_i)*	*f_ix_i*
0−10	6	5	30
10−20	8	15	120
20−30	10	25	250
30−40	15	35	525
40−50	5	45	225
50−60	4	55	220
60−70	2	65	130
Total	∑ f_i = 50		∑ f_ix_i = 1500

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{1500}{50} = 30

Thus, mean of the given data is 30.

Now, to find the median let us put the data in the table given below:

Class	*Frequency (f_i)*	Cumulative frequency (cf)
0−10	6	6
10−20	8	14
20−30	10	24
30−40	15	39
40−50	5	44
50−60	4	48
60−70	2	50
Total	N = ∑ f_i = 50

Now, N = 50

\Rightarrow \frac{N}{2} = 25 .

The cumulative frequency just greater than 25 is 39, and the corresponding class is 30−40.

Thus, the median class is 30−40.

∴ l = 30, h = 10, N = 50, f = 15 and cf = 24.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 30 + (\frac{25 - 24}{15}) \times 10 = 30 + \frac{10}{15} = 30.67

Thus, the median is 30.67.

Answer:

The frequency distribution table of less than type is as follows:

Class interval (upper class limits)	Cumulative frequency (cf)
Less than 10	2
Less than 15	2 + 12 = 14
Less than 20	14 + 2 = 16
Less than 25	16 + 4 = 20
Less than 30	20 + 3 = 23
Less than 35	23 + 4 = 27
Less than 40	27 + 3 = 30

and the frequency distribution table of more than type is as follows:

Class interval (lower class limits)	Cumulative frequency (cf)
More than 5	28 + 2 = 30
More than 10	16 + 12 = 28
More than 15	14 + 2 = 16
More than 20	10 + 4 = 14
More than 25	7 + 3 = 10
More than 30	3 + 4 = 7
More than 35	3

Taking upper class limits and lower class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

Here both ogives intersect at point A whose x−coordinate is 17.5.

Thus, the median of the data is 17.5.

Answer:

The frequency distribution table of less than type is as follows:

Production yield (in kg/ha) (upper class limits)	Cumulative frequency (cf)
Less than 45	1
Less than 50	1 + 9 = 10
Less than 55	10 + 15 = 25
Less than 60	25 + 18 = 43
Less than 65	43 + 40 = 83
Less than 70	83 + 26 = 109
Less than 75	109 + 16 = 125
Less than 80	125 + 14 = 139
Less than 85	139 + 10 = 149

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

The frequency distribution table of more than type is as follows:

Production yield (in kg/ha) (lower class limits)	Cumulative frequency (cf)
More than 40	148 + 1 = 149
More than 45	139 + 9 = 148
More than 50	124 + 15 = 139
More than 55	106 + 18 = 124
More than 60	66 + 40 = 106
More than 65	40 + 26 = 66
More than 70	24 + 16 = 40
More than 75	10 + 14 = 24
More than 80	10

Taking lower class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:

Answer:

The frequency distribution into the continuous form is as follows:

Marks	10.5−15.5	15.5−20.5	20.5−25.5	25.5−30.5	30.5−35.5	35.5−40.5	40.5−45.5	45.5−50.5
Number of students	2	3	6	7	14	12	4	2

Here the maximum class frequency is 14, and the class corresponding to this frequency is 30.5−35.5. So, the modal class is 30.5−35.5.

Now,

Modal class = 30.5−35.5, lower limit (l) of modal class = 30.5, class size (h) = 5,

frequency (f₁) of the modal class = 14,

frequency (f₀) of class preceding the modal class = 7,

frequency (f₂) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:

Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h = 30.5 + (\frac{14 - 7}{28 - 7 - 12}) \times 5 = 30.5 + (\frac{35}{9}) = 34.38

Hence, the mode is 34.38.

Now, to find the mean let us put the data in the table given below:

Class	*Frequency (f_i)*	*Class mark (x_i)*	*f_ix_i*
10.5−15.5	2	13	26
15.5−20.5	3	18	54
20.5−25.5	6	23	138
25.5−30.5	7	28	196
30.5−35.5	14	33	462
35.5−40.5	12	38	456
40.5−45.5	4	43	172
45.5−50.5	2	48	96
Total	∑ f_i = 50		∑ f_ix_i = 1600

Mean = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{1600}{50} = 32

Thus, mean of the given data is 32.

Now, to find the median let us put the data in the table given below:

Class	*Frequency (f_i)*	Cumulative frequency (cf)
10.5−15.5	2	2
15.5−20.5	3	5
20.5−25.5	6	11
25.5−30.5	7	18
30.5−35.5	14	32
35.5−40.5	12	44
40.5−45.5	4	48
45.5−50.5	2	50
Total	N = ∑ f_i = 50

Now, N = 50

\Rightarrow \frac{N}{2} = 25 .

The cumulative frequency just greater than 25 is 32, and the corresponding class is 30.5−35.5.

Thus, the median class is 30.5−35.5.

∴ l = 30.5, h = 5, N = 50, f = 14 and cf = 18.

Now,

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h = 30.5 + (\frac{25 - 18}{14}) \times 5 = 30.5 + \frac{35}{14} = 33

Thus, the median is 33.

Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 18 - Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

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