Question

The lifetime of a certain lightbulb is a random variable with the expectation of 8000 hours and a standard deviation of 200 hours. What is the probability that the average lifetime of 500 of these light bulbs is less than 8250 hours?

Answer 1

100% probability that the average lifetime of 500 of these light bulbs is less than 8250 hours

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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 = (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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 8000, \sigma = 200, n = 500, s = (200)/(√(500)) = 8.94`

What is the probability that the average lifetime of 500 of these light bulbs is less than 8250 hours?

This is the pvalue of Z when X = 8250. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (8250 - 8000)/(8.94)`

`Z = 27.95`

`Z = 27.95` has a pvalue of 1.

100% probability that the average lifetime of 500 of these light bulbs is less than 8250 hours