Answer: D. none of the above
Explanation:
Let x = a random variable that denotes the hourly income of midwives.
As per given , we have


Also, the distribution is normal.
Then, the probability that NJ midwives earn more than $60 per hour will be :_
![P(x>60)=1-P(x<60)=1-P((x-\mu)/(\sigma)<(60-55)/(15))\\\\=1-P(z<0.33)\ \ [\because\ z=(x-\mu)/(\sigma)]\\\\=1-0.6293\ \ [\text{By z-table}]\\\\ =0.3707=37.07\%](https://img.qammunity.org/2021/formulas/mathematics/college/xm84zwa1cikcpewtik0qud33qnc1xdeev0.png)
Hence, the percentage of NJ midwives earn more than $60 per hour is approximately 37.07%.
Since , 37.07% is not given in any option.
So the correct answer to this question is "D.none of the above"