Question

A new car battery is sold with a two-year warranty whereby the owner gets the battery replaced free of cost if it breaks down during the warranty period. Suppose an auto store makes a net profit of $20 on batteries that stay trouble-free during the warranty period; it makes a net loss of $10 on batteries that break down. The life of batteries is known to be normally distributed with a mean and a standard deviation of 40 and 16 months, respectively.

Required:
a. What is the probability that a battery will break down during the warranty period?
b. What is the expected profit of the auto store on a battery?
c. What is the expected monthly profit on batteries if the auto store sells an average of 500 batteries a month?

Answer 1

Answer:

a) 0.1587 = 15.87% probability that a battery will break down during the warranty period.

b) The expected profit of the auto store on a battery is of $15.239.

c) The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.

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.

The life of batteries is known to be normally distributed with a mean and a standard deviation of 40 and 16 months, respectively.

This means that $\mu = 40, \sigma = 16$

a. What is the probability that a battery will break down during the warranty period?

Two year warranty, that is, 24 months. This probability is the pvalue of Z when X = 24. So

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

$Z = \frac{24 - 40}{16}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.1587 = 15.87% probability that a battery will break down during the warranty period.

b. What is the expected profit of the auto store on a battery?

1 - 0.1587 = 0.8413 probability of a profit of $20.

0.1587 probability of a loss of $10. So

$E = 0.8413*20 - 0.1587*10 = 15.239$

The expected profit of the auto store on a battery is of $15.239.

c. What is the expected monthly profit on batteries if the auto store sells an average of 500 batteries a month?

Multiplying the average for a battery by 500. So

15.239*500 = $7,619.50

The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.