Answer: 6.68%
Step-by-step explanation:
Given : The mean weekly income of electrical engineers in NJ is $1,200 with a standard deviation of $200.
i.e. [tex]\mu=1200[/tex] and [tex]\sigma=200[/tex]
Let x denotes the monthly income ( in dollars).
If normally distributed , then the probability of workers earn more than $1,500 :
[tex]P(x>1500)=1-P(\leq1500)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{1500-1200}{200})\\\\=1-P(z\leq1.5)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\ =1-0.9332\ \ [\text{By z-table}]\\\\=0.0668=6.68\%[/tex]
Therefore , the percentages of workers earn more than $1,500 = 6.68%