Rs Aggarwal 2018 for Class 10 Math Chapter 9 - Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

Question 1:

If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x. [CBSE 2014]

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.

Question 2:

If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?

Answer:

â€‹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.

Question 3:

Compute the mean of following data: [CBSE 2013]

Class	1−3	3−5	5−7	7−9
Frequency	12	22	27	19

Answer:

The given data is shown as follows:

Class	*Frequency (f_i)*	*Class mark (x_i)*	*f_ix_i*
1−3	12	2	24
3−5	22	4	88
5−7	27	6	162
7−9	19	8	152
Total	∑ f_i = 80		∑ f_ix_i = 426

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{426}{80} = 5.325

Thus, the mean of the following data is 5.325.

Question 4:

Find the mean, using direct method:

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	7	5	6	12	8	2

Answer:

Class

Frequency

(f_{i})

Mid Values $(x_{i})$

$(f_{i} � x_{i})$

0-10

7

5

35

10-20

5

15

75

20-30

6

25

150

30-40

12

35

420

40-50

8

45

360

50-60

2

55

110

�

$\sum_{f_{i}} = 40$

�

$\sum (f_{i} � x_{i}) = 1150$

� $∴ �Mean,� \overset{�}{x} = \frac{\sum (f_{i} � x_{i})}{\sum f_{i}} =� \frac{1150}{40} =28 .75 ∴ \overset{�}{x} = 28 .75$

Question 5:

Find the mean, using direct method:

Class	25-35	35-45	45-55	55-65	65-75
Frequency	6	10	8	12	4

Answer:

Class

Frequency $(f_{i})$

Mid values $(x_{i})$

$(f_{i} � x_{i})$

25-35

6

30

180

35-45

10

40

400

45-55

8

50

400

55-65

12

60

720

65-75

4

70

280

�

$\sum f_{i} = 40$

�

$\sum (f_{i} � x_{i}) = 1980$

� $∴ �Mean,� \overset{�}{x} = \frac{\sum (f_{i} � x_{i})}{\sum f_{i}} =� \frac{1980}{40} =49 .5 ∴ \overset{�}{x} = 49.5$

Question 6:

Find the mean, using direct method:

Class	0-100	100-200	200-300	300-400	400-500
Frequency	6	9	15	12	8

Answer:

Class

Frequency $(f_{i})$

Mid values $(x_{i})$

$(f_{i} � x_{i})$

0-100

6

50

300

100-200

9

150

1350

200-300

15

250

3750

300-400

12

350

4200

400-500

8

450

3600

�

$\sum f_{i} = 50$

�

$\sum (f_{i} � x_{i}) = 13200$

� $∴ �Mean,� \overset{�}{x} = \frac{\sum (f_{i} � x_{i})}{\sum f_{i}} =� \frac{13200}{50} =264 ∴ \overset{�}{x} = 264$

Question 7:

Using an appropriate method, find the mean of following frequency distribution:

Class interval	84−90	90−96	96−102	102−108	108−114	114−120
Frequency	8	10	16	23	12	11

Which method did you use, and why?

Answer:

The given data is shown as follows:

Class interval	*Frequency (f_i)*	*Class mark (x_i)*	*f_ix_i*
84−90	8	87	696
90−96	10	93	930
96−102	16	99	1584
102−108	23	105	2415
108−114	12	111	1332
114−120	11	117	1287
Total	∑ f_i = 80		∑ f_ix_i = 8244

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{8244}{80} = 103.05

â€‹Thus, the mean of the following data is 103.05 .

Question 8:

If the mean of the following frequency distribution is 24, find the value of p. [CBSE 2013]

Class	0−10	10−20	20−30	30−40	40−50
Frequency	3	4	p	3	2

Answer:

The given data is shown as follows:

Class	*Frequency (f_i)*	*Class mark (x_i)*	*f_ix_i*
0−10	3	5	15
10−20	4	15	60
20−30	p	25	25p
30−40	3	35	105
40−50	2	45	90
Total	∑ f_i = 12 + p		∑ f_ix_i = 270 + 25p

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} \Rightarrow 24 = \frac{270 + 25 p}{12 + p} \Rightarrow 24 (12 + p) = 270 + 25 p \Rightarrow 288 + 24 p = 270 + 25 p \Rightarrow 25 p - 24 p = 288 - 270 \Rightarrow p = 18

Hence, the value of p is 18.

Question 9:

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is â‚¹18, find the missing frequency f.

Daily pocket allowance (in â‚¹)	11−13	13−15	15−17	17−19	19−21	21−23	23−25
Number of children	7	6	9	13	f	5	4

Answer:

The given data is shown as follows:

Daily pocket allowance (in â‚¹)	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

â€‹Hence, the value of f is 20.

Question 10:

If the mean of the following frequency distribution is 54, find the value of p. [CBSE 2006C]

Class	0−20	20−40	40−60	60−80	80−100
Frequency	7	p	10	9	13

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

â€‹Thus, the value of p is 11.

Question 11:

The mean of the following data is 42, find the missing frequencies x and y if the sum of the frequencies is 100.

Class interval	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	7	10	x	13	y	10	14	9

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

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

Question 12:

The daily expenditure of 100 families are given below. Calculate f₁ and f₂ if the mean daily expenditure is â‚¹188.

Expenditure (in â‚¹)	140−160	160−180	180−200	200−220	220−240
Number of families	5	25	f₁	f₂	5

Answer:

The given data is shown as follows:

Expenditure (in â‚¹)	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

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

Question 13:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	7	f₁	12	f₂	8	5

Answer:

Class

Frequency

(f_{i})

Mid values $(x_{i})$

$(f_{i} \times x_{i})$

0-20

7

10

70

20-40

f

₁

30

30 f

₁

40-60

12

50

600

60-80

18- f

₁

70

1260-70 f₁

80-100

8

90

720

100-120

5

110

550

$\sum f_{i} = 50$

$\sum (f_{i} \times x_{i}) = 3200 - 40 f_{1}$

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

Question 14:

During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarised as follows:

Number of heartbeats per minute	65−68	68−71	71−74	74−77	77−80	80−83	83−86
Number of patients	2	4	3	8	7	4	2

Find the mean heartbeats per minute for these patients, choosing a suitable method.

Answer:

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

Number of heartbeats per minute	Number of patients *(f_i)*	*Class mark (x_i)*	*f_ix_i*
65−68	2	66.5	133
68−71	4	69.5	278
71−74	3	72.5	217.5
74−77	8	75.5	604
77−80	7	78.5	549.5
80−83	4	81.5	326
83−86	2	84.5	169
Total	∑ f_i = 30		∑ f_ix_i = 2277

The mean of given data is given by

\bar{x} = \frac{\sum_{i} f_{i} x_{i}}{\sum_{i} f_{i}} = \frac{2277}{30} = 75.9

â€‹Thus, the mean heartbeats per minute for these patients is 75.9.

Question 15:

Find the mean, using assumed-mean method:

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of students	12	18	27	20	17	6

Answer:

Class

Frequency $(f_{i})$

Mid values $(x_{i})$

Deviation $(d_{i})$

d_{i} = (x_{i} - 25)

$(f_{i} � d_{i})$

0-10

12

5

-20

-240

10-20

18

15

-10

-180

20-30

27

25=A

0

30-40

20

35

10

200

40-50

17

45

20

340

50-60

6

55

30

180

�

\sum f_{i} = 100

�

$\sum (f_{i} � d_{i}) = 300$

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

Question 16:

Find the mean, using assumed-mean method:

Class	100-120	120-140	140-160	160-180	180-200
Frequency	10	20	30	15	5

Answer:

Class

Frequency

(f_{i})

Mid values $(x_{i})$

Deviation $(d_{i})$
$d_{i} = (x_{i} - 150)$

$(f_{i} � d_{i})$

100-120

10

110

-40

-400

120-140

20

130

-20

-400

140-160

30

150=A

0

160-180

15

170

20

300

180-200

5

190

40

200

�

$\sum f_{i} = 80$

�

$\sum (f_{i} � d_{i}) = - 300$

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

Question 17:

Find the mean, using assumed-mean method:

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	20	35	52	44	38	31

Answer:

Class

Frequency $(f_{i})$

Mid Values $(x_{i})$

Deviation $(d_{i})$
$d_{i} = (x_{i} - 50)$

$(f_{i} � d_{i})$

0-20

20

10

-40

-800

20-40

35

30

-20

-700

40-60

52

50=A

0

60-80

44

70

20

880

80-100

38

90

40

1520

100-120

31

110

60

1860

�

$\sum f_{i} = 220$

�

$\sum (f_{i} � d_{i}) = 2760$

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

Question 18:

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

Literacy rate (%)	45−55	55−65	65−75	75−85	85−95
Number of cities	4	11	12	9	4

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

â€‹Thus, the mean literacy rate is 69.5%.

Question 19:

Find the mean of the following frequency distribution using step-deviation method.

Class	0−10	10−20	20−30	30−40	40−50
Frequency	7	10	15	8	10

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

â€‹Thus, the mean is 25.8.

Question 20:

Find the mean of the following data, using step-deviation method.

Class	5−15	15−25	25−35	35−45	45−55	55−65	65−75
Frequency	6	10	16	15	24	8	7

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

â€‹Thus, the mean is 40.81.

Question 21:

â€‹The weights of tea in 70 packets are shown in the following table:

Weight (in grams)	200−201	201−202	202−203	203−204	204−205	205−206
Number of packets	13	27	18	10	1	1

Find the mean weight of packets using step-deviation method. [CBSE 2013]

Answer:

â€‹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

â€‹Hence, the mean is 201.96 g.

Question 22:

Find the mean of the following frequency distribution using a suitable method.

Class	20−30	30−40	40−50	50−60	60−70
Frequency	25	40	42	33	10

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

â€‹Thus, the mean is 42.534.

Question 23:

In an annual examination, marks (out of 90) obtained by students of class X in mathematics are given below:

Marks obtained	0−15	15−30	30−45	45−60	60−75	75−90
Number of students	2	4	5	20	9	10

Find the mean marks.

Answer:

â€‹Let us choose a = 52.5, h = 15, then d_i = x_i − 52.5 and u_i = $\frac{x_{i} - 52.5}{15}$ .

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

Marks obtained	*Number of students (f_i)*	*Class mark (x_i)*	d_i = x_i − 37.5	u_i = $\frac{x_{i} - 52.5}{15}$	*f_iu_i*
0−15	2	7.5	−45	−3	−6
15−30	4	22.5	−30	−2	−8
30−45	5	37.5	−15	−1	−5
45−60	20	52.5	0	0	0
60−75	9	67.5	15	1	9
75−90	10	82.5	30	2	20
Total	∑ f_i = 50				∑ f_iu_i = 10

The mean of given data is given by

\bar{x} = a + (\frac{\sum_{i} f_{i} u_{i}}{\sum_{i} f_{i}}) \times h = 52.5 + (\frac{10}{50}) \times 15 = 52.5 + 3 = 55.5

â€‹Thus, the mean is 55.5.

Question 24:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Age (in years)	18-24	24-30	30-36	36-42	42-48	48-54
Number of workers	6	8	12	8	4	2

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

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

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

Question 25:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	500-520	520-540	540-560	560-580	580-600	600-620
Frequency	14	9	5	4	3	5

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

560-580

4

570

1

4

580-600

3

590

2

6

600-620

5

610

3

15

$\sum f_{i} = 40$

$\sum (f_{i} \times u_{i}) = - 12$

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

Question 26:

Find the mean age from the following frequency distribution:

Age (in years)	25-29	30-34	35-39	40-44	45-49	50-54	55-59
No. of person	4	14	22	16	6	5	3

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

44.5-49.5

6

47

1

6

49.5-54.5

5

52

2

10

54.5-59.5

3

57

3

9

$\sum f_{i} = 70$

$\sum (f_{i} \times u_{i}) = - 37$

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

Question 27:

The following table shows the age distribution of patients of malaria in a village during a particular month.

Age (in years)	5-14	15-24	25-34	35-44	45-54	55-64
No. of cases	6	11	21	23	14	5

Find the average age of the patients.

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

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

$\sum f_{i} = 80$

$\sum (f_{i} \times u_{i}) = 43$

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

Question 28:

Weight of 60 eggs were recorded as given below:

Weight (in grams)	75−79	80−84	85−89	90−94	95−99	100−104	105−109
Number of eggs	4	9	13	17	12	3	2

Calculate their mean weight to the nearest gram.

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

â€‹

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

Question 29:

â€‹The following table shows the marks scored by 80 students in an examination:

Marks	Less than 5	Less than 10	Less than 15	Less than 20	Less than 25	Less than 30	Less than 35	Less than 40
Number of students	3	10	25	49	65	73	78	80

Calculate the mean marks correct to 2 decimal places.

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

â€‹

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

Question 1:

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. [CBSE 2014]

Age (in years)	0−15	15−30	30−45	45−60	60−75
Number of patients	5	20	40	50	25

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.

Question 2:

Compute the median from the following data:

Marks	0-7	7-14	14-21	21-28	28-35	35-42	42-49
Number of students	3	4	7	11	0	16	9

Answer:

Class

Frequency (f)

Cumulative

frequency

0-7

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

�

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

Question 3:

The following table shows the daily wages of workers in a factory:

Daily wages (in Rs)	0-100	100-200	200-300	300-400	400-500
Number of workers	40	32	48	22	8

Find the median daily wage income of the workers.

Answer:

Class

Frequency(f)

Cumulative

frequency

0-100

�

40

100-200

32

72

200-300

48

120

300-400

22

142

400-500

8

150

�

N=∑f=150

�

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

Question 4:

Calculate the median from the following frequency distribution:

Class	5-10	10-15	15-20	20-25	20-30	30-35	35-40	40-45
Frequency	5	6	15	10	5	4	2	2

Answer:

Class

Frequency(f)

Cumulative

frequency

5-10

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

�

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

Question 5:

Given below is the number of units of electricity consumed in a week in a certain locality:

Consumption (in units)	65-85	85-105	105-125	125-145	145-165	165-185	195-205
Number of consumers	4	5	13	20	14	7	4

Calculate the median

Answer:

Class

Frequency(f)

Cumulative

frequency

65-85

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

�

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

Question 6:

Calculate the median from the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
No. of boys	6	10	18	22	20	15	6	3

Answer:

Class

Frequency(f)

Cumulative

frequency

135-140

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

�

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

Question 7:

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	25	?	18	7

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

Question 8:

The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

Class	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	12	a	12	15	b	6	6	4

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.

Question 9:

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

Runs scored	2500−3500	3500−4500	4500−5500	5500−6500	6500−7500	7500−8500
Number of batsmen	5	x	y	12	6	2

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.

Question 10:

If the median of the following frequency distribution is 32.5, find the values of f₁ and f₂.

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	Total
Frequency	f₁	5	9	12	f₂	3	2	40

Answer:

Class

Frequency(f)

Cumulative

frequency

0-10

$f_{1}$

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

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

Question 11:

Calculate the median for the following data:

Age (in years)	19-25	26-32	33-39	40-46	47-53	54-60
Frequency	35	96	68	102	35	4

Answer:

First, we will convert the data into exclusive form.

Class

Frequency(f)

Cumulative

frequency

18.5-25.5

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

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

Question 12:

Find the median wages for the following frequencies distribution:

Wages per day (in Rs)	61-70	71-80	81-90	91-100	101-110	111-120
No. of women workers	5	15	20	30	20	8

Answer:

Converting the given data into exclusive form, we get:

Rs Aggarwal 2018 for Class 10 Math Chapter 9 - Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

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