Question

A company manufactures light bulbs. The lifetime for these bulbs is 4,000 hours with a standard deviation of 200 hrs. What lifetime should the company promote for these bulbs, whereby only 2% burnout before the claimed lifetime? (Hint: you may need to use norm.ppf() )

Answer 1

Answer:

The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime

Step-by-step 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:

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

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:

$\mu = 4000, \sigma = 200$

What lifetime should the company promote for these bulbs, whereby only 2% burnout before the claimed lifetime?

This is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.

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

$-2.055 = \frac{X - 4000}{200}$

$X - 4000 = -2.055*200$

$X = 3589$

The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime