I have a $ \mu = -1 $ and $ \theta = 6 $. I am supposed to find the probability $P(5 < X < 11)$.
My Attempt:
$$P(5< X < 11) $$ $$P(\frac{5--1}{6} < \frac{X--1}{6} < \frac{11--1}{6}) $$ $$ P(1 < Z < 2) $$
Now what? How do I calculate the Z-score without a table? Do I just integrate the normal probability density function (I tried that... I'm not sure if I am doing it wrong or what)?
btw, the answer the professor gave was:
$$P(1 < Z < 2) = .1359 $$
Given a CDF $F_X$, $P(a< X \leq b)=F(b)-F(a)$. You'll need a table or a calculator or something. Trying to integrate the PDF of the standard normal distribution will only bring you sadness.