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• RD Sharma Class 10 Solutions

# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 2

### Problem 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:

We know that, Mean = ∑fx/ N = 2470/1000 = 2.47

So, the mean number of heads per toss are 2.47

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

Solution:

Given,

Mean = 50, N = 120

We know that,

Mean = ∑fx/ N = (30f1 + 70f2 + 3480)/(68 + f1 + f2)

Now,

50 = (30f1 + 70f2 + 3480)/(68 + f1 + f2)

Also , 68 + f1 + f2 = 120

f1 = 52 – f2     ……. (i)

Now,

50 = (30f1 + 70f2 + 3480)/ 120

30f1 + 70f2 = 6000 – 3480

Now, putting the value of f1 from equation (i) –

30(52 – f2) + 70f2 = 2520

1560 – 30f2 + 70f2 = 2520

40f2 = 960

So, f2 = 24

and f1 = 52 – f2 = 52 – 24 = 28

Thus, f1 = 28 and f2 = 24

### Problem 13: The arithmetic mean of the following data is 14, find the value of k.

Solution:

Given,

Mean = 14

We know that,

Mean = ∑fx/ N = (360 + 10k)/(k + 24)

Now,

14(k + 24) = 360 + 10k

14k + 336 = 360 + 10k

4k = 24

So, k = 6

### Problem 14: The arithmetic mean of the following data is 25, find the value of k.

Solution:

Given,

Mean = 25

We know that,

Mean = ∑fx/ N = (390 + 15k)/(k + 14)

Now,

25(k + 14) = 390 + 15k

25k + 350 = 390 + 15k

10k = 40

So, k = 4

### Problem 15: If the mean of the following data is 18.75. Find the value of p.

Solution:

Given,

Mean = 18.75

We know that,

Mean = ∑fx/ N = (460 + 7p)/ 32

Now,

18.75 × 32 = 460 + 7p

600 = 460 + 7p

7p = 140

So, p = 20

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