Chapter 24: Measures of Central Tendency Exercise – 24.2

Question: 1

Calculate the mean for the following distribution:

Solution:

= 281/40 = 7.025.

Question: 2

Find the mean of the following data:

Solution:

= 2650/106 = 25.

Question: 3

The mean of the following data is 20.6 .Find the value of p.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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:

Solution:

= 800/40 = 20.

Question: 10

Candidates of four schools appear in a mathematics test. The data were as follows:

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

Solution:

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.

Solution:

∴ 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.

Solution:

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