Answer
a) .552
b) .9994
c) Part a
Work Step by Step
a. We use the z-scores to find:
$z=\frac{140-165}{45.6}=-.5482$
$z=\frac{211-165}{45.6}=1.01$
Thus, using the table of z-scores, we find that this corresponds to a probability of $.8438-.2912=..552$
b. We use the z-score to find:
$z=\frac{140-165}{45.6/\sqrt{36}}=-3.29$
$z=\frac{211-165}{45.6/\sqrt{36}}=6.06$
Thus, using the table of z-scores, we find that this corresponds to a probability of $.9999-.0005=.9994$
c. Part a is the most useful. After all, there is only one pilot, so it mainly matters how likely an individual woman is this weight.
