Question

In one region, the September energy consumption levels for single-family homes are normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh.
a. 0.1971
b. 0.3791
c. 0.2881
d. 0.0910
e. 0.8029

Answer 1

Answer:

a. 0.1971

Step-by-step explanation:

Z score is used to measure by how many standard deviations the raw score is above or below the mean. It is given by the formula:

$z=\frac{x-\mu}{\sigma}\\ \\Where\ \mu=mean, x=raw\ score, \sigma=standard\ deviation$

Given that:

μ = 1050 kWh, σ = 218 kWh

For x = 1100 kWh

$z=\frac{x-\mu}{\sigma}=\frac{1100-1050}{218} =0.23$

For x = 1225 kWh

$z=\frac{x-\mu}{\sigma}=\frac{1225-1050}{218} =0.80$

From the normal distribution table, P(1100 < x < 1225) = P(0.23 < z < 0.8) = P(z < 0.8) - P(z < 0.23) = 0.7881 - 0.5910 = 0.1971