Question

Each year, a large warehouse uses thousands of fluorescent light bulbs that burn 24 hours per day until they burn out and are replaced. The lifetime of the bulbs is a normally distributed random variable with mean equal to 600 hours and standard deviation of 12 hours. The supplier of the light bulbs and the manager agree that any bulb whose lifetime is among the lowest 1% of all possible lifetimes will be replaced at no charge. What is the maximum lifetime a bulb can have (hrs) and still be among the lowest 1% of all lifetimes

Answer 1

Answer:

The maximum lifetime a bulb can have and still be among the lowest 1% of all lifetimes is 572 hours.

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.

The lifetime of the bulbs is a normally distributed random variable with mean equal to 600 hours and standard deviation of 12 hours.

This means that $\mu = 600, \sigma = 12$

What is the maximum lifetime a bulb can have (hrs) and still be among the lowest 1% of all lifetimes?

This is the 1st percentile of lifetimes, which is X when Z has a pvalue of 0.01, that is, X when Z = -2.327.

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

$-2.327 = \frac{X - 600}{12}$

$X - 600 = -2.327*12$

$X = 572$

The maximum lifetime a bulb can have and still be among the lowest 1% of all lifetimes is 572 hours.