Let X and Y be normal random variables with means 0 and 2, respectively, and variances 1 and 9, respectively. Find the following, using the standard normal table. Express your answers to an accuracy of 4 decimal places.

P(X>0.75)=
P(X≤−1.25)=
Let Z=(Y−3)/4. Find the mean and the variance of Z.

E[Z]=
var(Z)=
P(−1≤Y≤2)=

4 answers

First two questions and last.

Z = (score-mean)/SD

SD^2 = variance

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score.
P(X>0.75)= 0.2266
P(X≤−1.25)= 0.1056

Can't figure out these:
E[Z]=
var(Z)=
P(−1≤Y≤2)=

anyone?
P(X>0.75)= 0.2266
P(X≤−1.25)= 0.1056
E[Z]= -1/4
var(Z)= 9/16
P(−1≤Y≤2)= ?
P(−1≤Y≤2)= 0.3413