Example 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49)...

Extracted text: Example 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49) x Pr{Y = 20} 21) %3D + (21 – 21.49)? x Pr{Y + (22 – 21.49)? x Pr{Y = 22} + (23 – 21.49) x Pr{Y = 23} = (-1.49) x 0.03 + (-.49) x 0.51 + (0.51) x 0.40 + (1.51)² × 0.06 = 2.2201 x 0.03 + .2401 x 0.51 + .2601 x 0.40 + 2.2801 x 0.06 = 0.066603 + 0.122451 + 0.10404 + 0.136806 = 0.4299. The standard deviation of Y is ay = V0.4299 - 0.6557. %3D