Posted by abc on .
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)=

Math (Partial) 
PsyDAG,
First two questions and last.
Z = (scoremean)/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. 
Math 
abc,
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? 
Math 
Bane Havoc,
P(X>0.75)= 0.2266
P(X≤−1.25)= 0.1056
E[Z]= 1/4
var(Z)= 9/16
P(−1≤Y≤2)= ? 
Math 
RVE,
P(−1≤Y≤2)= 0.3413