Answer: D. none of the above
Step-by-step explanation:
Let x = a random variable that denotes the hourly income of midwives.
As per given , we have
[tex]\mu=\$55[/tex]
[tex]\sigma=\$15[/tex]
Also, the distribution is normal.
Then, the probability that NJ midwives earn more than $60 per hour will be :_
[tex]P(x>60)=1-P(x<60)=1-P(\dfrac{x-\mu}{\sigma}<\dfrac{60-55}{15})\\\\=1-P(z<0.33)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.6293\ \ [\text{By z-table}]\\\\ =0.3707=37.07\% [/tex]
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"
