Chapter 7: Statistics Exercise – 7.1

Question: 1

Calculate the mean for the following distribution:

x:	5	6	7	9	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	281

Mean = 281/40 =7.025 

 

Question: 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
18	13	247
21	15	315
23	16	368
25	18	450
27	16	432
29	15	435
31	13	403
	N = 106	Sum = 2620

Mean (x) = 2680/106 = 25

Question: 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
5	10	150
P	25	25p
25	7	175
35	5	175
	N = 50	Sum = 2620 + 25P

Given Mean = 20.6

20.6 = (530 + 25p)/50

20.6 25p = 20.6 P

P = 20 

 

Question: 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	Sum = 10p + 445

Given Mean =15

15P + 405 = 10P + 445

15P - 10P = 445-405

5P = 40

P = 5

 

Question: 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	12	192
15	20	300
P	24	24p
20	16	320
25	8	200
30	4	120
	N = 100	Sum = 24p + 1228

Given Mean = 16.6

16.6 = (24p +1228)/100

1660  =  24p + 1228 

24p = 1660 – 1228

P = 432/24

P = 18 

 

Question: 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	24	480
25	2	50
	N = 50	Sum = 524 + 7p

Given mean = 12.58 
Mean = Sum/N
12.58 =  (524 +7p)/50
629 = 524 + 7p 
629 - 524 = 7p 
105 = 7P
P = 105/7
P = 15  

 

Question: 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 = P + 41	Sum =  9p + 303

Given Mean = 7.68

7.68 (P + 41) = 9P + 303
7.68 P + 314.88 = 9P + 303
9P – 7.68 P = 314.88 – 303
1.32 P = 11.88
P = 11.88/1.32
P = 9

 

Question: 8

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.

Ages (in years):	15	16	17	18	19	20
No of students:	3	8	10	10	5	4

Solution:

x	f	fx
15	3	45
16	8	128
17	10	170
18	10	180
19	5	95
20	4	80
	N = 40	Sum = 698

Mean age = sum/ N 
= 698/ 40 
= 17.45 years 

 

Question: 9

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	P	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 candidates from school III = P

Schools	No of candidates Ni	Average scores (xi)
I	60	75
II	48	80
III	P	55
IV	40	50

Given Average score or all schools = 66

10340 + 55p = 66 p + 9768 
10340 - 9768 = 66p –55 p 
P = 572/11 
P = 52

 

Question: 10

Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No of heads per toss	No of tosses
0	38
1	144
2	342
3	287
4	164
5	25
Total	1000

Solution:

No of heads per toss	No of tosses
0	38
1	144
2	342
3	287
4	164
5	25
Total	1000

 

No of heads per toss	No of tosses	fx
0	38	0
1	144	144
2	342	684
3	287	861
4	164	656
5	25	125

Mean number of heads per toss = 2470/1000 
= 2.47 
Mean = 2.47

 

Question: 11

The arithmetic mean of the following data is 25. Find the value of k.

x	5	15	25	35	45
f	3	k	3	6	2

Solution:

x	f	fx
5	3	15
15	k	15k
25	3	75
35	6	210
45	2	90
	N = k + 14	Sum = 15k + 390

Given mean = 25
Sum/ N = 25

25K + 350 = 15K + 390
25 K – 15K = 390 - 350
10 k = 40
K = 40/10
K = 4

 

Question: 12

If the mean of the following data is 18.75. Find the value of p.

x	10	15	p	25	30
f	5	10	7	8	2

Solution:

x	f	fx
10	5	50
15	10	150
20	p	20p
25	8	200
30	2	60
	N = P + 25	Sum = 20p + 460

Given mean = 18.75

18.75p + 468.75 = 20p + 460
468.75 – 460 = 20p – 18.75p
8.75 = 1.25P
P = 8.75/1.25
P = 7

 

Question: 13

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`	Sum = 295 +100p +5p2

Given Mean = 20

100p + 300 = 295 + 100p + 5p2
300 – 295 = 5p2 + 100p 
5 = 5p2 + 100P
5p2 + 100P – 5 =0
5p2 + 20p –5P
5(p2 – 1)  = 0
P2 – 1 = 0
P (+1, -1)
If p + 1 = 0, p = -1, Or p – 1 = 0, P = 1

 

Question: 14

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50. If ∑f = 120

x	10	30	50	70	90
f	17	f1	32	f2	19

Solution:

x	f	fx
10	17	170
30	f1	30f1
50	32	1600
70	f2	70f2
90	19	1710
	N = 68 + f1 + f2	Sum = 30f1 + 70f2 + 3480

Given mean = Sum/N
= 50

3400 + 50 f₁ + 50 f₂ = 30 f₁ + 70 f₂ + 3480
50 f₁ – 30 f₁ + 50 f₂ – 70 f₂ = 3480 – 3400
20 f₁ – 20 f₂ = 80
f₁ – f₂ = 4    ….. (i)
f₁ + f₂ + 68 = 120
f₁ + f₂ = 52 …. (ii)
Add both equation
f₁ = 28
f₂ = 24