Question

A salesman who uses his car extensively finds that his gasoline bills average $125.32 per month with a standard deviation of $49.51. Assume monthly gasoline bill amounts are normally distributed. The probability that his bill will be less than $50 a month or more than $150 for a single month is:

Answer 1

Answer:

The probability that his bill will be less than $50 a month or more than $150 for a single month is 0.3728 = 37.28%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

A salesman who uses his car extensively finds that his gasoline bills average $125.32 per month with a standard deviation of $49.51.

This means that $\mu = 125.32, \sigma = 49.51$

Less than 50:

p-value of Z when X = 50. So

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

$Z = \frac{50 - 125.32}{49.51}$

$Z = -1.52$

$Z = -1.52$ has a p-value of 0.0643

More than 150

1 subtracted by the p-value of Z when X = 150. So

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

$Z = \frac{150 - 125.32}{49.51}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915

1 - 0.6915 = 0.3085

The probability that his bill will be less than $50 a month or more than $150 for a single month is:

0.0643 + 0.3085 = 0.3728

The probability that his bill will be less than $50 a month or more than $150 for a single month is 0.3728 = 37.28%.