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

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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.

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If you can get away from always thinking in terms of $p$ values and $z$ scores and R commands and want to learn how to use tables of normal distribution probabilities (instead of just being told that the answers can be looked up in tables), this answer on math.SE might be useful. –  Dilip Sarwate Oct 6 '12 at 21:28

## marked as duplicate by whuber♦Oct 7 '12 at 19:23

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

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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 $$P(Z<\frac{m-a}{b})$$

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