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

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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 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
  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: 10 30 50 70 90  
f: 17 f1 32 f2 19 Total = 120

Solution:

x f fx
10 17 170
30 f1 30f1
50 32 1600
70 f2 70f2
90 19 1710
                         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 : 5 10 15 20 15
fi : 7 k 8 4 5

Solution:

x f fx
5 7 35
10 k 10k
15 8 120
20 4 80
25 5 125
                  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 : 5 15 25 35 45
fi : 3 k 3 6 2

Solution: 

x f fx
5 3 15
15 k 15k
25 3 75
35 6 210
45 2 90
                    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 : 10 15 p 25 30
fi : 5 10 7 8 2

Solution: 

x f fx
10 5 50
15 10 150
p 7 7p
25 8 200
30 2 60
                  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


Last Updated : 25 Jan, 2021
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