Answer to Question #176951 in Statistics and Probability for Sean

Question

Suppose that a random variable X is normally distributed with a mean of 200 and a variance of 

625.

Required: 

a) What proportion of X-values lies between 180 and 195? (7) 

b) Below what value do 33 per cent of X-values lie? (6)

c) What proportion of X-values are more than 190? (6)

d) What proportion of X-values are less than 195? (6)

~END~

Answer 1

a)

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

$\sigma=\sqrt{625}=25$

$z_1=\frac{180-200}{25}=-0.8$

$z_2=\frac{195-200}{25}=-0.2$

$P(180<X<195)=P(z<-0.2)-P(z<-0.8)=0.4207-0.2119=0.2088$

b)

$P(z<-0.44)=0.33=33\%$

$z=\frac{X-200}{25}=-0.44$

$X=-0.44\cdot25+200=189$

c)

$z=\frac{190-200}{25}=-0.4$

$P(X>190)=1-P(z<-0.4)=1-0.3446=0.6554$

d)

$z=\frac{195-200}{25}=-0.2$

$P(X<195)=P(z<-0.2)=0.4207$