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

Problem 1: Calculate the mean for the following distribution:

x:	5	6	7	8	9
f:	4	8	14	11	3

Solution:

x	f	fx
5	4	20
6	8	48
7	14	98
8	11	88
9	3	27
	N = 40	∑ fx = 281

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

Problem 2: Find the mean of the following data :

x:	19	21	23	25	27	29	31
f:	13	15	16	18	16	15	13

Solution: 

x	f	fx
19	13	247
21	15	315
23	16	368
25	18	450
27	16	432
29	15	435
31	13	403
	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:	10	15	p	25	35
f:	3	10	25	7	5

Solution:

x	f	fx
10	3	30
15	10	150
p	25	25p
25	7	175
35	5	175
	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:	5	10	15	20	25
f:	6	p	6	10	5

Solution: 

x	f	fx
5	6	30
10	p	10p
15	6	90
20	10	200
25	5	125
	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:	8	12	15	p	20	25	30
f:	12	16	20	24	16	8	4

Solution: 

x	f	fx
8	12	96
12	16	192
15	20	300
p	24	24p
20	16	320
25	8	200
30	4	120
	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:	5	8	10	12	p	20	25
f:	2	5	8	22	7	4	2

Solution:

x	f	fx
5	2	10
8	5	40
10	8	80
12	22	264
p	7	7p
20	4	80
25	2	50
	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:	3	5	7	9	11	13
f:	6	8	15	p	8	4

Solution:

x	f	fx
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	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:	15	17	19	20 + p	23
f:	2	3	4	5p	6

Solution: 

x	f	fx
15	2	30
17	3	51
19	4	76
20 + p	5p	100p + 5p2
23	6	138
	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):	15	16	17	18	19	20
No. of students:	3	8	10	10	5	4

Solution:

Age (in years)          (x)	No. of students            (f)	fx
15	3	45
16	8	128
17	10	170
18	10	180
19	5	95
20	4	80
	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 :

Schools	No. of Candidates	Average Score
I	60	75
II	48	80
III	NA	55
IV	40	50

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
I	60	75	4500
II	48	80	3840
III	p	55	55p
IV	40	50	2000
	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