Question

Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $109.72. Assume the amounts are normally distributed with standard deviation $24.00. Use the TI-84 Plus calculator to answer the following. (a) What proportion of bills are greater than $131

Answer 1

Answer:

18.67% of bills are greater than $131

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 109.72, \sigma = 24$

What proportion of bills are greater than $131

This proportion is 1 subtracted by the pvalue of Z when X = 131. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{131 - 109.72}{24}$

$Z = 0.89$

$Z = 0.89$ has a pvalue of 0.8133

1 - 0.8133 = 0.1867

18.67% of bills are greater than $131