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Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 2

  • Last Updated : 03 Mar, 2021
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Question 11. An incomplete distribution is given below :

Variable10-2020-3030-4040-5050-6060-7070-80
Frequency1230652518

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up missing frequencies.

(ii) Calculate the AM of the completed distribution.

Solution:

Let us assume p1, and p2 to be the missing frequencies

Median = 46 and N = 230



VariableFrequency (f)cf
10-201212
20-303042
30-40p142+p1
40-5065107+p1
50-60p2107+p1+p2
60-7025132+p1+p2
70-8018150+p1+p2
Total 230 

∴ 150 + p1 + p2 = 230

⇒ p1+p2 = 230 – 150 = 80

∴ p2 = 80-p1 …..(i)

Since, median = 46 which lies in the class interval belonging to 40-50

∴ I = 40, f= 65, F = 42 +p1, h = 10

Median = l + (\frac{\frac{N}{2}-F}{f})*h \\ 46 = 40 + \frac{\frac{230}{2} - (42 + p1)}{65}*10 \\ 6 = \frac{10}{65}(\frac{230}{2} - 84 - 2p1) \\ 6 = \frac{2}{13}(\frac{146-2p1}{2}) \\ \frac{78}{2} = \frac{146 - 2p1}{2}

⇒ 39 = 73 – p1

⇒ p1 = 73 -39 = 34



∴ p2 – 80 – p1 = 80 – 34 = 46

Therefore, the missing frequencies are 34, and 46.

Let the assumed mean (A) be 45.

VariableClass Marks (x)Frequency (f)

d = x -A 

A = 45

fi * di
10-201512-30-360
20-302530-20-600
30-403534-10-340
40-5045 – A6500
50-60554610460
60-70652520500
70-80751830540
Total 230 200

Mean, A = \frac{\sum f_id_i}{\sum f_i} \\ = 45 + \frac{200}{230}

= 45 + 0.8695

= 45 + 0.87

= 45.87

Question 12. If the median of the following frequency distribution is 28.5 find the missing frequencies:

Class interval0-1020-3030-4040-5050-60Total
Frequency5f115f2560

Solution:



Mean = 28.5, N = 60

Class intervalFrequencyc.f
0-1055
10-20f15 +f
20-302025 + f1
30-401540 + f1
40-50f240 + f1 + f2
50-60545 + f1 + f2
Total 60 

Therefore, 

45 + f1 + f2 = 60

=> f1 + f2 = 60 – 45 = 15

=> f2 = 15 – f1

17 = 25 – f1

N/2 = 30 

Now, Median = 28.5 and it lies in the class interval of 20-30

Therefore, 

l = 20, F = 5 + f1, f= 20 and h = 10

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ => 28.5 = 20 + \frac{30 - (5+f_1)}{20} * 10 \\ => 28.5 = 20 + \frac{30-5-f_1}{2} \\ => 8.5 * 2 = 25 - f_1

⇒ f1= 25 -17 = 8

and f2 = 15-f1 = 15-8 = 7

Therefore, the missing frequencies are 8 and 7 respectively. 

Question 13. The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Class IntervalFrequencyClass IntervalFrequency
0-1002500-60020
100-2005600-700f2
200-300f1700-8009
300-40012800-9007
400-50017900-10004

Solution:

Median = 525, N = 100

Class IntervalFrequencyc.f.
0-10022
100-20057
200-300f17 + f1
300-4001219 + f1
400-5001736 + f1
500-6002056 + f1
600-700f256 + f1 + f2
700-800965 + f1 + f2
800-900772 + f1 + f2
900-1000476 + f1 + f2
Total100 

Therefore,

76 + f1 + f2 = 100 => f1 + f2 = 100 – 76 = 26

f = 24 – f1

Because, 



Median = 525 which belongs to the interval 500-600

Now, l =500, F = 36 + f1, f =20, h = 100

Therefore, 

MMedian = l + \frac{\frac{N}{2}-F}{f} * h \\ = 500 + \frac{50 - (36 + f_1)}{20} * 100 \\ => 525 = 500 + \frac{50-36-f_1}{20} * 100

⇒ 525 – 500 = (14 -f1) x 5

⇒ 25 = 70- 5f1

⇒ 5f1 = 70 – 25 = 45

⇒ f1 = 455 = 9

and f2 = 24 – f1 = 24 – 9 = 15

Hence, we obtain the values for f1 = 9, f2 = 15.

Question 14. If the median of the following data is 32.5, find the missing frequencies.

Class interval0-1010-2020-3030-4040-5050-6060-70Total
Frequencyf15912f23240

Solution:

Mean = 32.5 and N= 40

Class intervalFrequency (f)c.f.
0-10f1f1
10-2055 + f1
20-30914 + f1
30-401226 + f1
40-50f226 + f1 + f2
50-60329 + f1 + f2
60-70231 + f1 + f2
 40 

Now, we know, 

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ 32.5 = 30 + \frac{\frac{40}{2}-(14+f_1)}{12} * 10 \\ 32.5 - 30 = \frac{20-14-f_1}{12} *10 \\ 2.5 = \frac{6-f_1}{12}*10

Solving, we get,

⇒ 2.5 x 12 = 60 – 10f1

⇒ 30 = 60 – 10f1

⇒ 10f1 = 60-30 = 30

⇒ f1 = 30/10 =3

∴ f2 = 9 – f1 = 9-3 = 6



Hence, f1 = 3, f2= 6

Question 15. Compute the median for each of the following data:

(i) 

MarksNo. of students(ii) MarksNo. of students
Less than 100More than 1500
Less than 3010More than 14012
Less than 5025More than 13027
Less than 7043More than 12060
Less than 9065More than 110105
Less than 11087More than 100124
Less than 13096More than 90141
Less than 150100More than 80150

Solution:

(i) Less than

Marksc.ff
0-1000
10-301010
30-502515
50-704318
70-906522
90-1108722
110-130969
130-1501004

We have, N= 100

∴N/2 = 100/2 = 50 which lies in the class  interval belonging to 70-90 (∵ 50 < 65 and > 43)

∴ l = 70, F =43 , f = 22 ,h = 20

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ = 70 + \frac{50-43}{22} * 20 \\ = 70 +  \frac{7 * 20}{22}   \\ = 70 + \frac{70}{11} \\ = 76.36

(ii) Greater than

Marksc.ff
More than 150 (150-160)00
140-1501212
130-1402715
120-1306035
110-12010545
100-11012419
90-10014117
80-901509

We have,



N = 150, N/2 = 150/2 = 75 which lies in the class interval belonging to 110-120 (∵ 75 > 105 and 75 > 60)

∴ l = 110, F = 60 , f=45, h= 10

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ = 110 + \frac{70-60}{45} * 10 \\ = 110 +  \frac{15}{45} * 10   \\ = 110 + \frac{10}{3} \\ = 113.33

Question 16. A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained.

Height in cmNumber of girls
Less than 1404
Less than 14511
Less than 15029
Less than 15540
Less than 16046
Less than 16551

Find the median height.

Solution:

Height (in cm)No of girls (c.f)F
135 – 14044
140 – 145117
145 – 1502918
150 – 1554011
155 – 160466
160 – 165515
  51

Here, ∑F/2 = 51/2 = 25.5 or 26 which lies in the class interval belonging to 145-150

Therefore,

l= 145, F= 11, f= 18, h= 5

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ 145 + \frac{25.5-11}{18} * 5 \\ 145 +  \frac{14.5}{18} * 5 \\ 145 + \frac{72.5}{18}

= 145 + 4.03 = 149.03

Question 17. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

Age in yearsNumber of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:



Age in years

No of policy holders

(c.f)

f
15-2022
20-2564
25-302418
30-354521
35-407833
40-458911
45-50923
50-55986
55-601002
Total 100

Here N = 100, N/2 = 100/2 = 50 which lies in the class interval of 35-40 ( ∵ 45 > 50> 78)

Therefore, 

l = 35, F = 45, f= 33, h = 5

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ 35 + \frac{50-45}{33} * 5 \\ 35 +  \frac{5}{33} * 5 \\ 35 + \frac{25}{33}

= 35 + 0.76 = 35.76

Question 18. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm)118-126127-135136-144145-153154-162163-171172-180
No of leaves35912542

Find the mean length of leaf.

Solution:

Length (in mm)

(in exclusive form)

No. of leaves (f)c.f.
117.5 – 126.533
126.5 – 135.558
135.5 – 144.5917
144.5 – 153.51229
153.5 – 162.5534
162.5 – 171.5438
171.5 – 180.5240

N = 40, N/2 = 40/2 = 20 which lies in the class interval of 144.5-153.5 as 17 < 20 < 29

Therefore, 

l= 144.5, F= 17, f= 12, h = 9

Median = l + \frac{\frac{N}{2}-F}{f} * h \\ 144.5 + \frac{20-17}{12} * 9 \\ 144.5 +  \frac{3}{12} * 9 \\ 144.5 + \frac{9}{4}

= 144.5 + 2.25 = 146.75

Question 19. An incomplete distribution is given as follows :

Variable0-1010-2020-3030-4040-5050-6060-70
Frequency1020?40?2515

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Solution:

Median = 25 and ∑f= N = 170

Let us assume x and y to be the two missing frequencies

VariableFrequencyc.f
0-101010
10-202030
20-30x30+x
30-404070+x
40-50y70+x+y
50-602595+x+y
60-7015110+x+y

∴ 110 + x +y = 170

⇒ x + y = 170 – 110 = 60



Here, we have,

 N = 170, N/2 = 170/2 = 85

Therefore, Median = 35 which lies in the class interval belonging to 30-40

Here l = 30, f= 40, F = 30 + x and h = 10

Median = l + \frac{\frac{n}{2}-F}{f} * h \\ 35 = 30 + \frac{85-(30+x)}{40} * 10 \\ 35 - 30 =  \frac{85-30-x}{4} \\ 5 = \frac{55-x}{4}

20 = 55 – x

⇒ x = 55 – 20 = 35

But,

x + y = 60

Solving for y, we get,

∴ y = 60 – x = 60 – 35 = 25

Hence missing frequencies x and y are 35 and 25.

Question 20. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Class interval0-66-1212-1818-2424-30
Frequency4x5y1

Solution:

Class intervalFrequencyCumulative Frequency
0-644
6-12x4+x
12-1859+x
18-24y9+x+y
24-30110+x+y

We know, n = 20

Therefore, 

10 + x + y – 20, 

=> x+y= 10 …(i)

Also,

Median = 14.4 which lies in the class interval belonging to 12-18

So, l = 12, f= 5, cf = 4 + x, h = 6

Median = l + \frac{\frac{n}{2}-cf}{f} * h \\ 14.4 = 12 + \frac{10-(4+x)}{5} * 6 \\ 14.4 = 12 + (\frac{6-x}{5})*6

Solving for x, we get, 

x = 6 ….(ii)

Also, 

y = 6 

Question 21. The median of the following data is SO. Find the values of p and q, if the sum of all the frequencies is 90.

Marks:20-3030-4040-5050-6060-7070-8080-90
Frequency:p152520q810

Solution: 

MarksFrequencyCumulative Frequency
20-30pp
30-401515 + p 
40-502540 + p = cf
50-6020 = f60 + p
60-70q60 + p + q
70-8068 + p + q
80-901078 + p + q

Given, N = 90

And, N/2 = 90/2 = 45 which lies in the class interval 50-60

Now, 

Lower limit, l = 50, f= 20, cf= 40 + p, h = 10



Median = l + \frac{\frac{N}{2}-cf}{f} * h \\ = 50 + \frac{45-40-p}{20} * 10 \\ 50 = 50 + (\frac{5-p}{2}) \\ 0 = \frac{5-p}{2}

Obtaining values, we get, 

∴ P = 5

Also, 78 +p + q = 90

⇒ 78 + 5 + q = 90

⇒ q = 90-83

∴ q = 7

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