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() )

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 and standard deviation , the zscore of a measure X is given by:

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:

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.

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

what are ur choices?

it’s the first one your welcome