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EXAMPLE 4.35 Find the Mean and the Variance

CASE 4.2 In Case 4.2 (pages 210211), 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 (xiμx)2ρi
0 0.202 0.00 (02.754)2(0.202)=1.53207
1 0.159 0.159 (12.754)2(0.159)=0.48917
2 0.201 0.402 (22.754)2(0.201)=0.11427
3 0.125 0.375 (32.754)2(0.125)=0.00756
4 0.088 0.352 (42.754)2(0.088)=0.13662
5 0.087 0.435 (52.754)2(0.087)=0.43887
6 0.056 0.336 (62.754)2(0.056)=0.59004
7 0.025 0.175 (72.754)2(0.025)=0.45071
8 0.022 0.176 (82.754)2(0.022)=0.60545
9 0.018 0.162 (92.754)2(0.018)=0.70223
10 0.008 0.080 (102.754)2(0.008)=0.42004
11 0.006 0.066 (112.754)2(0.006)=0.40798
12 0.003 0.036 (122.754)2(0.003)=0.25647
μX=2.754 σ2X=6.151
Page 230

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.

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