Question

The Boffo Product Company sells a waffle iron on which they have done product testing. They have determined that the amount of time the product will last can be described by a normal distribution. In particular, the average waffle iron lasts for 12 years and one standard deviation is 8 months. How long should they warranty the product for if they want no more than 6.7% of the waffle irons to fail within that time

Answer 1

Answer:

They should warranty the product for 7 years if they want no more than 6.7% of the waffle irons to fail within that time.

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 average waffle iron lasts for 12 years and one standard deviation is 8 months.

Measuring the time in months, we have that $\mu = 12*8 = 96$ and $\sigma = 8$

How long should they warranty the product for if they want no more than 6.7% of the waffle irons to fail within that time?

This is X when Z has a p-value of 0.067, so X when Z = -1.5. Then

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

$-1.5 = \frac{X - 96}{8}$

$X - 96 = -1.5*8$

$X = 84$

84 months = 7 years.

They should warranty the product for 7 years if they want no more than 6.7% of the waffle irons to fail within that time.