Answer to Question #152989 in Statistics and Probability for Aditya jaiswal

Question

If a random variable X follows Normal distribution such that P (9.65<=X<=13.8)=0.07008and P(X>9.6)=0.8159 where Find the mean & variance of X and find P(X>=13.8 /X>=9.6).

Answer 1

$P(X>9.6) = P(Z > \frac {9.6 - \mu} {\sigma}) =\\ 1 - P(Z \le \frac {9.6 - \mu} {\sigma}) = 0.8159\\ \frac {9.6 - \mu} {\sigma}=Z_{1-0.8159}=Z_{0.1841} =-0.8999$

$P(9.6\le X\le13.8) = P(X\ge 9.6) - P(X\ge 13.8) = 0.07008\\ P(X \ge 13.8) = 0.8159 - 0.07008 = 0.74582\\ \frac {13.8 - \mu} {\sigma}=Z_{1-0.74582}=Z_{0.25418} =-0.6614$

From two equations we found that sigma = 17.6133, mean = mu = 25.4493,

var = s²=310.227

And

$P(X\ge13.8 |X\ge 9.6) =\frac{P(X\ge13.8)}{P(X\ge 9.6)}=\frac{0.74582}{0.8159}= 0.9141$