Question

The average electric bill for a small electric company is $72 for the month of April with a standard deviation of $6. If the amounts of the bills are normally distributed, find the probability that the mean of the bills for 15 randomly selected residents will be less than $75

Answer 1

Answer:

Hence the probability that the mean of their utility bills will be less than $75 = P(X < 75) is 0.9736.

Step-by-step explanation:

Now,  

Population mean electric bill, $\mu =$  $72  

Population standard deviation, $\sigma =$  $6  

Standard error of the mean$= \sigma / \sqrt{n} = 6 / \sqrt{15} = 1.5492$

The probability that the mean of their utility bills will be less than $75 = P(X < 75)

= P[Z < (75 - 72) / 1.5492]

= P[Z < 1.9365]

= 0.9736 (Using Z table)