Answer to Question #128524 in Statistics and Probability for Gazal

Question

If X is distributed normally with mean 12 and standard deviation 4. Find i) P(X>20) , ii) P(X<20), iii)

p(0<X<12)

Answer 1

$\mu = 12$

$\sigma = 4$

Solution i) p(X>20)

$p(X>20) = 1 - p(X<20)$

$Z = \frac{x - \mu}{\sigma}$ = $\frac{20-12}{4}$ = 2

$p(X<20)$ = probability at Z = 2

= 0.9772

$p(X > 20) = 1 - 0.9772$

Answer: 0.0228

Solution ii) p(X<20)

From (i) p(X<20) = 0.9772

= Probability at Z = 2

Answer: 0.9772

solution iii) p(0<X<12)

$p(0<X<12)$ =

$p(x<12) - p(x<0)$

For p(X<12)

at

$Z = \frac{12-12}{4} = 0$

p(x<12) = 0.5

For p(x<0)

at

$Z=\frac{0-12}{4} = -3$

p(x<0) = 0.00135

$P(0<x<12) = 0.5 - 0.00135$

Answer: 0.49865