Given a normal distribution, a mean, and standard deviation what is the probability a Variable is in a range [duplicate]

Question

Possible Duplicate:
Normal distribution probability

Issues getting to the bottom of a HW problem, but I am not looking for the answer, just some guidance.

x has a normal distribution with the specified mean = 15.9 and standard deviation = 3.6

Find the indicated probability P(10 ≤ x ≤ 26)?

I assume i need to take the values 10 and 26 and calculate Z-Scores, and then go the Z-Score table and take the differences in the P-Values? Is this the right approach or am I missing something? I followed that reasoning and the answer was wrong - so either I made a rounding mistake, or have chosen the wrong path.

Thank you for the advice in advance.

Answer 1

Am I missing something or is this basic probability calculus? for any r.v. $X$ with a known cumulative distribution function $F_X (x)$, $P(a<x\leq b) = F_X (b) - F_X (a)$.

Answer 2

You need to perform transformation of the random variable with distribution $$ X \sim \mathcal{N}(a=15.9,b^2=3.6^2) $$ to normal variable $$ Z \sim \mathcal{N}(0,1) $$

To do this use the formula: $$ P(X<m) = P( \frac{X-a}{b} < \frac{m-a}{b})=P(Z<\frac{m-a}{b}) $$

The idea is that the expression $$Z = \frac{X-a}{b} \sim \mathcal{N}(0,1)$$ has standard normal distribution. Then you can use std. distribution table to get values of $$P(Z<\frac{m-a}{b})$$

Answer 3

I redid the problem using R (I didn't know the pnorm function existed) as opposed to the Z-sore table and the process was correct (the correct answer was one hundredth off, so I must have made a typo). I suppose I just assumed I did the problem wrong.