We are given the mean `bar(x)=68` and standard deviation `s=4.2`

(a) Find the proportion of females whose height is greater than 69 inches.

First, convert the raw score of 69 to a `z` score:

`z=(x-bar(x))/s=(69-68)/4.2=.2381` (Note that the `z` score is positive; this indicates that the data value is above the mean)

Since the population is approximately normal, we can consider the area under the standard normal curve. The proportion under the curve to the right of `z=.2381` is the proportion greater than 69 in. From the standard normal table we find .5941 or 59.41% is to the left, so .4059 or 40.59% is to the right.(There are different representations of the standard normal table; some give the area to the left of a given z score, others the area from the mean to the z score. I used a graphing calculator which agrees with the table value.)

**Approximately 40.6% of the population is taller than 69 in.**

(b) The proprtion no greater than 69 in is the complement of those greater than 69in, so **59.4% are no more than 69 in tall.**

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