A salesman who uses his car extensively finds that his gasoline bills average $125.32 per month with a standard deviation of $49.51.The probability that his bill will be less than $50 a month or more than $150 for a single month is:
A) 0.4357.
B) 0.1915.
C) 0.6272.
D) 0.3728.
E) 0.6915.

Question

Quizplus · Accepted Answer

To solve this problem, we need to standardize the values using z-score formula:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the population mean, and σ is the population standard deviation.

For the lower limit of $50, we have:

z = (50 - 125.32) / 49.51 = -1.521

For the upper limit of $150, we have:

z = (150 - 125.32) / 49.51 = 0.497

Now we can calculate the probabilities using the standard normal distribution table or calculator:

P(z < -1.521) + P(z > 0.497) = 0.0682 + 0.3046 = 0.3728

Therefore, the probability that his bill will be less than $50 a month or more than $150 for a single month is 0.3728, which corresponds to choice D.

A Salesman Who Uses His Car Extensively Finds That His