Question

In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at the manufacturer's expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense? (Round z-score computation to 2 decimal places. Round your final answers to 2 decimal places.)

Answer 1

Answer:

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. This means that $\mu = 36.84, \sigma = 3.34$.

Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense?

This is the value of X when Z has a pvalue of 0.90.

Looking at the z-table, we get that this is between $Z = 1.28$ and $Z = 1.29$, so we use $Z = 1.285$.

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

$1.28 = \frac{X - 36.84}{3.34}$

$X - 36.84 = 1.28*3.34$

$X = 41.12$

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.