EXAMPLE 4.35 Find the Mean and the Variance
CASE 4.2 In Case 4.2 (pages 210–211), we saw that the distribution of the daily demand X of transfusion blood bags is
Bags used | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
Probability | 0.202 | 0.159 | 0.201 | 0.125 | 0.088 | 0.087 | 0.056 |
Bags used | 7 | 8 | 9 | 10 | 11 | 12 | |
Probability | 0.025 | 0.022 | 0.018 | 0.008 | 0.006 | 0.003 |
We can find the mean and variance of X by arranging the calculation in the form of a table. Both μX and σ2X are sums of columns in this table.
xi | pi | xipi | (xi−μx)2ρi |
---|---|---|---|
0 | 0.202 | 0.00 | (0−2.754)2(0.202)=1.53207 |
1 | 0.159 | 0.159 | (1−2.754)2(0.159)=0.48917 |
2 | 0.201 | 0.402 | (2−2.754)2(0.201)=0.11427 |
3 | 0.125 | 0.375 | (3−2.754)2(0.125)=0.00756 |
4 | 0.088 | 0.352 | (4−2.754)2(0.088)=0.13662 |
5 | 0.087 | 0.435 | (5−2.754)2(0.087)=0.43887 |
6 | 0.056 | 0.336 | (6−2.754)2(0.056)=0.59004 |
7 | 0.025 | 0.175 | (7−2.754)2(0.025)=0.45071 |
8 | 0.022 | 0.176 | (8−2.754)2(0.022)=0.60545 |
9 | 0.018 | 0.162 | (9−2.754)2(0.018)=0.70223 |
10 | 0.008 | 0.080 | (10−2.754)2(0.008)=0.42004 |
11 | 0.006 | 0.066 | (11−2.754)2(0.006)=0.40798 |
12 | 0.003 | 0.036 | (12−2.754)2(0.003)=0.25647 |
μX=2.754 | σ2X=6.151 |
We see that σ2X=6.151. The standard deviation of X is σX=√6.151=2.48. The standard deviation is a measure of the variability of the daily demand of blood bags. As in the case of distributions for data, the connection of standard deviation to probability is easiest to understand for Normal distributions (for example, 68–95–99.7 rule). For general distributions, we are content to understand that the standard deviation provides us with a basic measure of variability.