The heights of males in a certain country are normally distributed with a mean of 70.2 inches and a standard deviation of 4.2 inches. What is the probability that a randomly chosen male is under the height of 65 inches?

Feb 14, 2017

Since 65 is (approx.) -1.24 standard deviations from the mean of 70.2,

P(X < 65)" "~~" "P(Z < "–"1.24)
color(white)(P(X < 65)" ")=" "0.1075.

Explanation:

If $X$ is $N \left(\mu , {\sigma}^{2}\right)$ and $Z$ is $N \left(0 , 1\right)$, then

$P \left(X < x\right) = P \left(Z < \frac{x - \mu}{\sigma}\right) .$

This means we can find the probability $P \left(X < x\right)$ by computing the corresponding probability from the standard normal distribution $Z$.

Let $X$ be the height of a random male from the given country. Then $X \text{ "~" } N \left(70.2 , 4.2\right)$, and so

$P \left(X < 65\right) = P \left(Z < \frac{65 - 70.2}{4.2}\right)$
$\textcolor{w h i t e}{P \left(X < 65\right)} \approx P \left(Z < \text{–} 1.24\right)$

What this means is that the point in the $X$ distribution where $X = 65$ corresponds (approximately) to the point in the $Z$ distribution where $Z = \text{–} 1.24$. In other words, 65 is (approx.) 1.24 standard deviations below the mean of 70.2.

Why bother "transforming" $X$ like this? Because it would be impossible to print a look-up table of $x$-scores for every possible normal distribution, but we can print one table as a base, then show how to relate any normal random variable to this base one.

The table we so happen to make is for $Z$, the standard normal distribution. Thus, looking up $P \left(Z < \text{–} 1.24\right)$ in our $z$-score table is identical to looking up $P \left(X < 65\right)$ in its table (if it existed).

From the table, $P \left(Z < \text{–} 1.24\right) = 0.1075$, and so

$P \left(X < 65\right) \approx 0.1075$.