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

  • Last Updated : 25 Jan, 2021
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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.

No. of heads per tossNo. of tosses           
038
1144
2342
3287
4164
525
Total1000

Solution: 

No. of heads per toss

               (x)

No. of tosses

        (f) 



fx
0380
1144144
2342684
3287861
4164656
525125
 Total (N) = 1000   ∑ fx = 2470      

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.

x:1030507090 
f:17f132f219Total = 120

Solution:

xffx
1017170
30f130f1
50321600
70f270f2
90191710
                        N = 68 + f1 + f2 = 120 ∑ fx = 30f1 + 70f2 + 3480  

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.

xi :510152015
fi :7k845

Solution:

xffx
5735
10k10k
158120
20480
255125
                 N = k + 24  ∑ fx = 360 + 10k

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.

xi :515253545
fi :3k362

Solution: 

xffx
5315
15k15k
25375
356210
45290
                   N = 14 + k  ∑ fx = 390 + 15k

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.

xi :1015p2530
fi :510782

Solution: 

xffx
10550
1510150
p77p
258200
30260
                 N = 32      ∑ fx = 460 + 7p

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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