Question

It has a time to failure distribution which is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles. Find its designed life if a .97 reliability is desired.

Answer 1

Answer:

The designed life should be of 21,840 vehicle miles.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.

This means that $\mu = 35000, \sigma = 7000$

Find its designed life if a .97 reliability is desired.

The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.

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

$-1.88 = \frac{X - 35000}{7000}$

$X - 35000 = -1.88*7000$

$X = 21840$

The designed life should be of 21,840 vehicle miles.