im not exactly sure, the question did not say that Z was a standard normal variable, only that is was normal. But the letter Z usually denotes a std normal variable...
Im not exactly sure how you got Pr(X>2k)=Pr(Z>1), could you clarify?
im not exactly sure, the question did not say that Z was a standard normal variable, only that is was normal. But the letter Z usually denotes a std normal variable...
Im not exactly sure how you got Pr(X>2k)=Pr(Z>1), could you clarify?
If X~N(k,k^2), then if we change to Z-scores:
z=(x-mean(x))/sd(x)=(x-k)/k
then X>2k is equivalent to X-k>k, which is equivalent to (X-k)/k>1.