Question 1182643
The distribution of X is approximately normal with mean $187,000 and variance 45000^2 $^2.
a. z=(x bar-mean)/sigma
>=-350/45000 which is >=-0.00778
<=(8500)/45000 or 0.18888
that probability is 0.0780 (unrounded, which z often is not, would be 0.0771
b .for 36 people the denominator would be 45000/sqrt(36) or 7500
This probability would be 0.3901 (not rounded).

The point of this question is that it would be quite unlikely for a single person to have a salary in this restricted area around the mean, but the mean of a group of 36 would be far more likely to cluster around the mean.