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Question
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were 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 1
| No. of heads per toss | No. of tosses | fx |
| 0 | 38 | 0 |
| 1 | 144 | 144 |
| 2 | 342 | 684 |
| 3 | 287 | 861 |
| 4 | 164 | 656 |
| 5 | 25 | 125 |
| N = 1000 | `sum`fx = 2470 |
Mean number of heads per toss `=(sumfx)/N`
`=2470/1000=2.47`
∴ Mean = 2.47
Solution 2
Let the assumed mean (A) = 2
| No. of heads per toss (x1) | No. of intervals (f1) |
u1 = x1 - A (A = 2) |
f1u1 |
| 0 | 38 | -2 | -76 |
| 1 | 144 | -1 | +44 |
| 2 | 342 | 0 | 0 |
| 3 | 287 | 1 | 287 |
| 4 | 164 | 2 | 328 |
| 5 | 25 | 3 | 75 |
| N = 1000 | `sumf_1u_1=470` |
Mean number of per toss `=A+(sumf_1u_1)/N`
`=2+470/1000`
= 2 + 0.47
= 2.47
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