Question

The average life of a certain type of small motor is 10 years with a standard deviation of 2 years. The manufacturer replaces free all motors that fail while under guarantee. If she is willing to replace 3% of the motors that fail, how long a guarantee (in years) should she offer? You may assume that the lives of the batteries are normally distributed. (Hint: you may need to use norm.ppf() )

Answer 1

Answer:

She should offer a guarantee of 6.24 years.

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 10, \sigma = 2$

If she is willing to replace 3% of the motors that fail, how long a guarantee (in years) should she offer?

This is the value of X when Z has a pvalue of 0.03. So it is X when Z = -1.88. So

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

$-1.88 = \frac{X - 10}{2}$

$X - 10 = -1.88*2$

$X = 6.24$

She should offer a guarantee of 6.24 years.