Question

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1075 kWh.

Answer 1

Solution :

Given that ,

mean = = 1050

standard deviation = = 218

n = 50

= 1050 and

= / n = 218 / 50 = 30.8299

P( > 1075) = 1 - P( < 1075)

= 1 - P(( - ) / < (1075 - 1050) / 30.8299)

= 1 - P(z < 0.81)

= 1 - 0.791

= 0.209

Probability = 0.209