Answer
a. .6517
b. .9115
c. It should be reduced.
Work Step by Step
a. We use the z-score to find:
$z=\frac{167-182.9}{40.8}=-.39$
Thus, using the table of z-scores, we find that this corresponds to a probability of $1-.3483=.6517$
b. We use the z-score to find:
$z=\frac{167-182.9}{40.8/\sqrt{12}}=-1.35$
Thus, using the table of z-scores, we find that this corresponds to a probability of $1-.0885=.9115$
c. For safety standards, this probability is too low, so the weight limit should be reduced.
