If X is distributed as $N(\mu,\sigma)$, then the standardized variable $\displaystyle Z=\frac{X-\mu}{\sigma}$ has the standard normal distribution. This property of the normal distribution allows us to cast a probability problem concerning X into one concerning Z. To find the probability that X lies in a given interval, convert the interval to the z-scale and then calculate the probability by using the standard normal table.

If X is distributed as $N(\mu,\sigma)$, then

$\begin{array}{rcl} \displaystyle P[a\leq X\leq b]&=&P[\displaystyle\frac{a-\mu}... ...frac{b-\mu}{\sigma}]-P[\displaystyle Z\leq\frac{a-\mu}{\sigma}] \\ \end{array}$

where Z has the standard normal distribution.

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The number of calories in a salad on the lunch menu is normally distributed with mean=200 and sd=5. Find the probability that the salad you select will contain
(a)
More than 208 calories.
(b)
Between 190 and 200 calories.

Letting X denote the number of calories in the salad, we have the standardized variable $\displaystyle Z=\frac{X-200}{5}$

(a)
The z-value corresponding to x=208 is $\displaystyle z=\frac{208-200}{5}=1.6$. Therefore,
$\begin{array}{rcl} P[X>208]&=&P[Z>1.6] \\ &=&1-P[Z\leq 1.6] \\ &=&1-.9452=.0548 \\ \end{array}$
(b)
The z-values corresponding to x=190 and x=200 are $\frac{190-200}{5}=-2.0$ and $\frac{200-200}{5}=0$ respectively. We calculate
$\begin{array}{rcl} P[190\leq X\leq 200]&=&P[-2.0\leq Z\leq 0] \\ &=&.5-.0228=.4772 \\ \end{array}$