Chapter 24: Measures of Central Tendency Exercise – 24.2

Question: 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

= 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
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

= 2650/106 = 25.

 

Question: 3

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 = 25p + 530

It is given that,

Mean = 20.6

⇒ 25p + 530 = 20.6 × 50

⇒ 25p = 1030 − 530

⇒ 25p = 500

⇒ p = 500/25 = 20

⇒ p = 20

∴ p = 20.

 

Question: 4

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

f:	6	p	6	10	5
x:	5	10	15	20	25

Solution:

x	f	fx
5	6	30
10	P	10p
15	6	90
20	10	200
25	5	125
	N = p + 27	∑fx = 10p + 445

It is given that,

Mean = 15

⇒∑fx/N = 15

⇒10p + 445p + 27=15

⇒10p + 445 = 15 × (p + 27)

⇒ 10p + 445 = 15p + 405

⇒15p − 10p = 445 − 405

⇒ 5p = 40

⇒ p = 405 = 8

⇒ p = 8

∴ p = 8.

 

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	16	192
15	20	300
P	24	24p
20	16	320
25	8	200
30	4	120
	N = 100	∑fx = 24p + 1228

It is given that,

Mean = 16.6

⇒ 24p + 1228 = 1660

⇒ 24p = 1660 − 1228

⇒ 24p = 432

⇒ p = 432/24 = 18

⇒ p = 18

∴ 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	4	80
25	2	50
	N = 50	∑fx = 7p + 524

It is given that,

Mean = 12.58

⇒ ∑fx/N = 12.58

⇒ 7p + 524 = 629

⇒ 7p = 629 − 524

⇒ 7p = 105

⇒ p = 1057 = 15

⇒ p = 15

∴ p = 18.

 

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	∑fx = 9p + 303

It is given that,

Mean = 7.68

⇒ ∑fx/N = 7.68

⇒ 9p + 303p + 41 = 7.68

⇒ 9p + 303 = 7.68p + 314.88

⇒ 9p − 7.68p = 314.88 − 303

⇒ 1.32p = 11.88

⇒ p = 11.881.32 = 9

⇒ p = 9

∴ p = 9.

 

Question: 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 = 5p2 + 100p + 295

It is given that,

Mean = 20

⇒ 5p² + 100p + 295 = 20(5p + 15)

⇒ 5p² + 100p + 295 = 100p + 300

⇒ 5p² = 300 − 295

⇒ 5p² = 5

⇒ p² = 1

⇒ p = ±1

Frequency can’t be negative.

Hence, value of p is 1.

 

Question: 9

Find the mean of the following distribution:

x:	10	12	20	25	35
f:	3	10	15	7	5

Solution:

x	f	fx
10	3	30
12	10	120
20	15	300
25	7	175
35	5	175
	N = 40	∑fx = 800

= 800/40 = 20.

 

Question: 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	Not Available	55
IV	40	50

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

Solution:

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

Given the average score of all schools = 66

⇒ 10340 + 55x = 66x + 9768

⇒ 10340 − 9768 = 66x − 55x

⇒ 11x = 572

⇒ x = 572/11 = 52

∴ No. of candidates appeared from school III = 52.

 

Question: 11

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 (x)	No. of tosses (f)	fx
0	38	0
1	144	144
2	342	684
3	287	861
4	164	656
5	25	125
	N = 1000	∑fx = 2470

∴ Mean number of heads per toss = ∑fx/N

= 2470/1000

= 2.47

 

Question: 12

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

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 = 120	∑fx = 3480 + 30f1 + 70f2

It is given that

Mean = 50

⇒ ∑fx/N = 50

⇒ 3480 + 30f₁ + 70f₂ = 50 × 120

⇒30f₁ + 70f₂ = 6000 − 3480

⇒10(3f₁ + 7f₂) = 10(252)

⇒ 3f₁ + 7f₂ = 252 ⋅⋅⋅⋅ (1)  [∵ Divide by 10]

And N = 20

⇒ 17 + f₁ + 32 + f₂ + 19 = 120

⇒ 68 + f₁ + f₂ = 120

⇒ f₁ + f₂ = 120 − 68

⇒ f₁ + f₂ = 52

Multiply with 3 on both sides

⇒ 3f₁ + 3f₂ = 156 ⋅⋅⋅ (2)

Subtracting equation (2) from equation (1)

⇒ 3f₁ + 7f₂ − 3f₁ − 3f₂ = 252 − 156

⇒ 4f₂ = 96

⇒ f₂ = 96/4 = 24

Put the value of f₂ in equation (1)

⇒ 3f₁ + 7 × 24 = 252

⇒ 3f₁ = 252 − 168

⇒ f₁ = 84/3 = 28

⇒ f₁ = 28