Answer to Question #258068 in Statistics and Probability for Busi

Question

A public utility noted that the monthly electricity usage (measured in kilowatt-hours) by certain residential users is normally distributed with a mean of 1040. If the monthly electricity usage of a sample of 15 randomly selected users yielded a variance of 6500, what is the probability that their mean electrical usage exceeds 1100?

Answer 1

$$P(X>1100)=1-P(X\leq1100)$$

$$=1-P(Z\leq \dfrac{1100-\mu}{s/\sqrt{n}})$$

$$=1-P(Z\leq \dfrac{1100-1040}{6500/\sqrt{15}})$$

$$\approx1-P(Z\leq0.03575)=0.485741$$

The probability that their mean electrical usage exceeds 1100 is $0.485741.$