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. for a randomly selected home, find the probability that the september energy consumption level is between 1100 kwh and 1225 kwh. round to four decimal places.

a. 0.2881
b. 0.3791
c. 0.1982
d. 0.0910

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1050\), \(\sigma=218\).
For \(x = 1100\):
\(z_1=\frac{1100 - 1050}{218}=\frac{50}{218}\approx0.23\)
For \(x = 1225\):
\(z_2=\frac{1225-1050}{218}=\frac{175}{218}\approx0.80\)

Step2: Use the standard normal distribution table

We want to find \(P(0.23<Z<0.80)\).
We know that \(P(0.23 < Z<0.80)=P(Z < 0.80)-P(Z < 0.23)\)
From the standard normal distribution table, \(P(Z < 0.80)=0.7881\) and \(P(Z < 0.23)=0.5910\)
\(P(0.23 < Z<0.80)=0.7881 - 0.5910=0.1971\approx0.1982\) (due to rounding differences in table values)

Answer:

C. 0.1982