EXAMPLE 4.34
Find the mean and the variance. In Example 4.32 (pages 254–255) , we saw that the distribution of the number X of fall courses taken by students at a small liberal arts college is
| Courses in the fall | 1 | 2 | 3 | 4 | 5 | 6 |
| Probability | 0.05 | 0.05 | 0.13 | 0.26 | 0.36 | 0.15 |
We can find the mean and variance of X by arranging the calculation in the form of a table. Both μX and are sums of columns in this table.
| xi | pi | xi pi | (xi − μX)2pi |
|---|---|---|---|
| 1 | 0.05 | 0.05 | (1 − 4.28)2(0.05) = 0.53792 |
| 2 | 0.05 | 0.10 | (2 − 4.28)2(0.05) = 0.25992 |
| 3 | 0.13 | 0.39 | (3 − 4.28)2 (0.13) = 0.21299 |
| 4 | 0.26 | 1.04 | (4 − 4.28)2(0.26) = 0.02038 |
| 5 | 0.36 | 1.80 | (5 − 4.28)2(0.36) = 0.18662 |
| 6 | 0.15 | 0.90 | (6 − 4.28)2(0.15) = 0.44376 |
| μX = 4.28 | = 1.662 | ||
We see that = 1.662. The standard deviation of X is . The standard deviation is a measure of the variability of the number of fall courses taken by the students at the small liberal arts college. As in the case of distributions for data, the standard deviation of a probability distribution is easiest to understand for Normal distributions.