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

Answer 1

Answer:

She should offer a guarantee of 13.76 years.

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 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.

The average life of a certain type of small motor is 10 years with a standard deviation of 2 years.

This means 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?

She should offer the 100 - 3 = 97th percentile as a guarantee, so X when Z has a pvalue of 0.97, that is, X when Z = 1.88.

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

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

$X - 10 = 2*1.88$

$X = 13.76$

She should offer a guarantee of 13.76 years.