normal distribution question

Quote:

Originally Posted by djmccabie

ok not sure what im doin wrong with question hope someone can help.

The randow variable X has a normal distribution with mean 16 and variance 0.64. Find x, such that P(X<x) = 0.025

this is how i attempt to solve it -

X~N(16,0.8²)

(X-16)/0.8=0.025

X=16.02

but i be doin it wrong me think :/

any suggestions??

thanks

$X\sim N(16,0.8^2)$

so: $Z=(X-16)/0.8 \sim N(0,1)$

The value of $z$ such that:

$Pr(Z<z)=0.025$

is obtained by reverse look up in a table of the standard normal distribution, and that tells us that:

$Pr(Z<1.96)=0.025$

$Z<1.96$ becomes:

$X-16<1.96 \times 0.8$

or:

$X<16+ 1.96 \times 0.8$

CB