Question

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?

Answer 1

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(mu, sigma^2)` and `Z` is `N(0,1)`, then

`P(X < x)= P(Z< (x-mu)/sigma).`

This means we can find the probability `P(X < x)` 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" "~" "N(70.2,4.2)`, and so

`P(X < 65)=P(Z < (65-70.2)/4.2)`
`color(white)(P(X < 65)) ~~ P(Z < "–"1.24)`

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="–"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(Z < "–"1.24)` in our `z`-score table is identical to looking up `P(X < 65)` in its table (if it existed).

From the table, `P(Z < "–"1.24)=0.1075`, and so

`P(X < 65) ~~ 0.1075`.