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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 1

  • Last Updated : 25 Jan, 2021

Problem 1: Calculate the mean for the following distribution:

x:56789
f:814113

Solution:

ffx
5420
6848
1498
81188
9327
            N = 40 ∑ fx = 281

We know that, Mean = ∑fx/ N = 281/40 = 7.025

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Problem 2: Find the mean of the following data :

x:19212325272931
f:13151618161513

Solution: 



ffx
1913247
2115315
2316368
2518450
2716432
2915435
3113403
              N = 106 ∑ fx = 2650

We know that, Mean = ∑fx/ N = 2650/106 = 25

Problem 3: If the mean of the following data is 20.6. Find the value of p.

x:1015p2535
f:3102575

Solution:

xffx
10330
1510150
p2525p
257175
355175
             N = 50 ∑ fx = 530 + 25p

Given,

Mean = 20.6

We know that,

Mean = ∑fx/ N = (530 + 25p)/50

Now,

20.6 = (530 + 25p)/ 50



(20.6 × 50) – 530 = 25p

p = 500/25

So, p = 20

Problem 4: If the mean of the following data is 15, find p.

x:510152025
f:6p6105

Solution: 

xffx
5630
10p10p
15690
2010200
255125
                 N = p + 27 ∑ fx = 445 + 10p

Given,

Mean = 15

We know that,

Mean = ∑fx/ N = (445 + 10p)/(p + 27)

Now,

15 = (445 + 10p)/(p + 27)



15p + 405 = 445 + 10p

5p = 40

So, p = 8

Problem 5: Find the value of p for the following distribution whose mean is 16.6

x:81215p202530
f:121620241684

Solution: 

xffx
81296
1216192
1520300
p2424p
2016320
258200
304120
                N = 100 ∑ fx = 1228 + 24p

Given,

Mean = 16.6

We know that,

Mean = ∑fx/ N = (1228 + 24p)/ 100

Now,

16.6 = (1228 + 24p)/ 100



1660 = 1228 + 24p

24p = 432

So, p = 18

Problem 6: Find the missing value of p for the following distribution whose mean is 12.58

x:581012p2025
f:25822742

Solution:

xffx
5210
8540
10880
1222264
p77p
20480
25250
              N = 50 ∑ fx = 524 + 7p

Given,

Mean = 12.58

We know that,

Mean = ∑fx/ N = (524 + 7p)/ 50

Now, 

12.58 = (524 + 7p)/ 50

629 = 524 + 7p

7p = 105

So, p = 15

Problem 7: Find the missing frequency (p) for the following distribution whose mean is 7.68

x: 35791113
f:6815p84

Solution:

xffx
3618
5840
715105
9p9p
11888
13452
                  N = 41 + p ∑ fx = 303 + 9p

Given,

Mean = 7.68

We know that,

Mean = ∑fx/ N = (303 + 9p)/(41 + p)

Now,

7.68 = (303 + 9p)/(41 + p)



7.68(41 + p) = 303 + 9p

7.68p + 314.88 = 303 + 9p

1.32p = 11.88

So, p = 9

Problem 8: Find the value of p, if the mean of the following distribution is 20.

x:15171920 + p23
f:2345p6

Solution: 

xffx
15230
17351
19476
20 + p5p100p + 5p2
236138
                     N = 5p + 15 ∑ fx = 295 + 100p + 5p2

Given,

Mean = 20

We know that,

Mean = ∑fx/ N = (295 + 100p + 5p2)/(5p + 15)

Now,

20 = (295 + 100p + 5p2)/(5p + 15)

20(5p + 15) = 295 + 100p + 5p2

100p + 300 = 295 + 100p + 5p2

5p2 – 5 = 0

5(p2 – 1) = 0

p2 – 1 = 0

(p + 1)(p – 1) = 0

So, p = 1 or -1

Here, p = -1 (Reject as frequency of a number cannot be negative)

So, p = 1

Problem 9: The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students.

Age (in years):151617181920
No. of students:38101054

Solution:

Age (in years)

         (x)

No. of students 

          (f)

fx
15345
168128
1710170
1810180
19595
20480
 N = 40∑ fx = 698

We know that, Mean = ∑fx/ N = 698/40 = 17.45

So, the mean age of the students is 17.45 years.

Problem 10: Candidates of four schools appear in a mathematics test. The data were as follows :

SchoolsNo. of CandidatesAverage Score
I6075
II4880
III NA55
IV               4050

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

Solution:

Let the number of candidates that appeared from school III = p

Schools

No. of Candidates  



             (f)

Average Score

         (x)

fx
I60754500
II48803840
IIIp5555p
IV                40502000           
                     N = 148 + p                         ∑ fx = 10340 + 55p 

Given,

Average score of the candidates of all the four schools = Mean = 66

We know that,

Mean = ∑fx/ N = (10340 + 55p)/(148 + p)

Now,

66 = (10340 + 55p)/(148 + p)

66(148 + p) = 10340 + 55p

66p + 9768 = 10340 + 55p

11p = 572

So, p = 52

So, the number of candidates that appeared from school III are 52




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