Answer to Question #152931 in Statistics and Probability for xyz

Question

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

Answer 1

From the equation;

P(X>9.6)=0.8159

P(x>z)=0.8159

z=-0.9

P(X>13.8) = P(X>9.6) - P (9.6$\leq$ X$\leq$ 13.8) = 0.8159 - 0.07008 = 0.7458

P(x>z)=0.7458

z=-0.661

z=$\frac{X-\mu}{\sigma}$

-0.9=$\frac{9.6-\mu}{\sigma}$

-0.661=$\frac{13.8-\mu}{\sigma}$

$-0.9*\sigma+\mu=9.6$

$-0.661*\sigma+\mu=13.8$

$\sigma=17.6$

$\mu=25.4$

P(X$\ge$13.8|X$\ge$9.6) = P(X$\ge$13.8)/P(X$\ge$9.6) = 0.7458/0.8159 = 0.914