RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1 . Calculate the mean for the following distribution :

SOLUTION :

Q 2 . Find the mean of the following data :

SOLUTION :

Q 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 = 50025 = 20

⇒ p = 20

∴ p = 20.

Q 4 . If the mean of the following data is 15 , find p.

SOLUTION :

It is given that ,

Mean = 15

⇒ 10p+445 = 15×(p+27)

⇒10p+445 = 15p+405

⇒15p−10p = 445−405

⇒ 5p = 40

⇒ p = 40/5 =8

⇒ p = 8

∴ p = 8.

Q 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.

Q 6 . Find the missing value of p for the following distribution whose mean is 12.58 .

SOLUTION :

It is given that ,

Mean = 12.58

⇒ 7p+524 = 629

⇒ 7p = 629−524

⇒ 7p = 105

⇒ p = 1057 = 15

⇒ p = 15

∴ p = 18.

Q7 . Find the missing frequency (p) for the following distribution whose mean is 7.68 .

SOLUTION :

It is given that ,

Mean = 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.

Q 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.

Q 9 . Find the mean of the following distribution :

SOLUTION :

Q 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.

Q.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

= 2470/1000

= 2.47

Q 12 . Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

Total = 120

SOLUTION :

It is given that

Mean = 50

⇒ 3480+30f₁+70f₂ = 50×120

⇒ 30f₁+70f₂ = 6000−3480

⇒10(3f₁+7f₂) = 10(252)

⇒3f₁+7f₂ = 252⋅⋅⋅⋅⋅(1)                         [∵Divideby10]

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