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 only 3% of the motors that fail, how long a guarantee should be offered? Assume that the lifetime of a motor follows a normal distribution.

Respuesta :

Answer:

A guarantee of 6.24 years should be offered.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 10, \sigma = 2[/tex]

If she is willing to replace only 3% of the motors that fail, how long a guarantee should be offered?

The 3rd percentile, that is, the value of X when Z has a pvalue of 0.03. So X when Z = -1.88.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.88 = \frac{X - 10}{2}[/tex]

[tex]X - 10 = -1.88*2[/tex]

[tex]X = 6.24[/tex]

A guarantee of 6.24 years should be offered.