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?

1 Answer
Feb 14, 2017

Answer:

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#.