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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() )

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Answer:

She should offer a guarantee of 6.24 years.

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 = (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 = (X - \mu)/(\sigma)


-1.88 = (X - 10)/(2)


X - 10 = -1.88*2


X = 6.24

She should offer a guarantee of 6.24 years.

User Hatjhie
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