Answer to Question #317715 in Statistics and Probability for Student2014_

Question

Suppose a certain population of observations are normally distributed. What percentage of the observations in the population are within 1.5 standard deviations of the mean?

Events A1 , A2, and A3 are mutually exclusive and exhaustive events, with probabilities P(A1) = 0.20, P(A2) = 0.60, and P(A3) = 0.20. Given that P(B|A1) = 0.10, P(B|A2) = 0.50, and P(B|A3) = 0.40, calculate P(A3|B).

Answers to 4 decimal points.

Answer 1

$P\left( \left| X-\mu \right|\leqslant 1.5\sigma \right) =P\left( \left| \frac{X-\mu}{\sigma} \right|\leqslant 1.5 \right) =P\left( \left| Z \right|\leqslant 1.5 \right) =\\=2\varPhi \left( 1.5 \right) -1=2\cdot 0.933-1=0.866\\\\P\left( A_3|B \right) =\frac{P\left( B|A_3 \right) P\left( A_3 \right)}{P\left( B|A_1 \right) P\left( A_1 \right) +P\left( B|A_2 \right) P\left( A_2 \right) +P\left( B|A_3 \right) P\left( A_3 \right)}=\\=\frac{0.4\cdot 0.2}{0.1\cdot 0.2+0.5\cdot 0.6+0.4\cdot 0.2}=0.2$