Ten coins were tossed together a number of times. The distribution of the number of heads obtained is given in the following table :
| No of heads |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
| Frequency |
2 |
7 |
23 |
36 |
11 |
61 |
100 |
12 |
8 |
5 |
3 |
Calculate, correct to three decimal places, the :
(a) mean number of heads ;
(b) probability of getting an even head ;
(c) probability of getting an odd number.
Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
| \(f\) | 2 | 7 | 23 | 36 | 11 | 61 | 100 | 12 | 8 | 5 | 3 | 268 |
| \(fx\) | 0 | 7 | 46 | 108 | 44 | 305 | 600 | 84 | 64 | 45 | 30 | 1333 |
(a) Mean number of heads.
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{1333}{268} = \mathbf{4.974} \]
(b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).
\[ P(\text{even}) = \frac{147}{268} = \mathbf{0.549} \]
(c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).
\[ P(\text{odd}) = \frac{121}{268} = \mathbf{0.451} \]
Check: \(0.549 + 0.451 = 1.000\).
Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
| \(f\) | 2 | 7 | 23 | 36 | 11 | 61 | 100 | 12 | 8 | 5 | 3 | 268 |
| \(fx\) | 0 | 7 | 46 | 108 | 44 | 305 | 600 | 84 | 64 | 45 | 30 | 1333 |
(a) Mean number of heads.
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{1333}{268} = \mathbf{4.974} \]
(b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).
\[ P(\text{even}) = \frac{147}{268} = \mathbf{0.549} \]
(c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).
\[ P(\text{odd}) = \frac{121}{268} = \mathbf{0.451} \]
Check: \(0.549 + 0.451 = 1.000\).