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 $30 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 standard deviation of 45 and 12 months respectively. a. What is the probability that the battery will break down during the warranty period

Answer 1

Answer:

4.01% probability that the battery will break down during the warranty period

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 45, \sigma = 12$

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

Warranty of 2 years = 24 months. So this is the pvalue of Z when X = 24.

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

$Z = \frac{24 - 45}{12}$

$Z = -1.75$

$Z = -1.75$ has a pvalue of 0.0401

4.01% probability that the battery will break down during the warranty period