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.

We are given that 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.

Also, 50 different homes are randomly selected.

Let [tex]\bar X[/tex] = sample mean energy consumption level

The z-score probability distribution for sample mean is given by ;

Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean energy consumption levels = 1050 kWh

[tex]\sigma[/tex] = standard deviation = 218 kWh

n = sample of homes = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that their mean energy consumption level for September is greater than 1075 kWh is given by = P([tex]\bar X[/tex] > 1075 kWh)

P([tex]\bar X[/tex] > 1075) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{1075-1050}{\frac{218}{\sqrt{50} } }[/tex] ) = P(Z > 0.81) = 1 - P(Z [tex]\leq[/tex] 0.81)

= 1 - 0.79103 = 0.20897

The above probability is calculated using z table by looking at value of x = 0.81 in the z table which have an area of 0.79103.

Therefore, probability that their mean energy consumption level for September is greater than 1075 kWh is 0.20897.