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.
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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