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# 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