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. 0.1982
b. 0.2881
c. 0.3791
d. 0.0810

Explanation:

Step1: Calculate 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: Find probabilities using standard normal table

We know that \(P(1100<X<1225)=P(Z < z_2)-P(Z < z_1)\)
From the standard normal table, \(P(Z < 0.80)=0.7881\) and \(P(Z < 0.23)=0.5910\)
\(P(0.23<Z<0.80)=0.7881 - 0.5910=0.1981\approx0.1982\)

Answer:

A. 0.1982