Thread: normal distribution question

Consider a normally distributed random variable where mu = std. dev = k.
Find Pr(Z > 2k):

No idea on this one.
(the answer is 0.1587 for reference)

Originally Posted by scorpion007

Consider a normally distributed random variable where mu = std. dev = k.
Find Pr(Z > 2k):

No idea on this one.
(the answer is 0.1587 for reference)

For clarification do we have:

X~N(k,k^2)

Z=(X-k)/k

find Pr(Z>2k)?

or:

X~N(k,k^2)

find Pr(X>2k)?

If the last of these is what is intended then set

Z=(X-k)/k,

then Z~N(0,1)

and

Pr(X>2k)=Pr(Z>1)

which can be looked up in a standard normal table.

RonL

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?

Originally Posted by scorpion007

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.

Now Z~N(0,1),

and Pr(Z>1)=0.1587

RonL