Applications

Example 1

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If a woman is randomly selected, find the probability that her height is between 61.6 inches and 68.6 inches.

Solution .

We need to find P(61.6<x<68.6). The area we need to find is graphed below.

To calculate this we first convert the endpoints to z scores using

61.6 corresponds to z = (61.6 - 63.6)/2.5 = (-2)/2.5 = -.8

68.6 corresponds to z = (68.6 - 63.6)/2.5 = (5)/2.5 = 2

The corresponding region on the z-scale is given below.

Thus, P(61.6<x<68.6) = P(-.8<z<2) = P(-.8<z<0) + P(0<z<2) = .2881 + .4772 = .7653

If we convert this to a percentage, we can say that about 76.53% of women are between 61.6 inches and 68.6 inches tall.

Example 2

Using the population from example one, find the probability that a woman selected at random is over 70 inches tall.

Solution .

We need to find P(x>70). The region whose area we need to find is sketched below.

To calculate this we first convert the endpoint to a z score using

70 corresponds to z = (70 - 63.6)/2.5 = (6.4)/2.5 = 2.56

Below is the corresponding graph on the z-scale.

P(x>70) = P(z>2.56) = .5 - P(0<z<2.56) = .5 - .4948 = .0052

Note that the computer generated answer given in the captions on the two graphs translates into .00523. Again the discrepancy is due to the fact that the computer used 10 digit values in its calculations rather the 4 digit values given in the tables.

Example 3

Of 10000 randomly selected women how many would you expect would be taller than 5 feet 10 inches tall?

Solution .

Note that 5 feet 10 inches is the same as 70 inches. In example two we saw that the probability of randomly selecting a woman who was taller than 70 inches was .0052. Multiply this by 10000 and we get 52. Thus we would expect that 52 women would be taller than 5 feet 10 inches.