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USE CODE: SELF10

Chapter 24: Measures of Central Tendency Exercise – 24.2

Quesiton: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 4

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

f: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 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.

Quesiton: 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

Quesiton: 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

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