Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 6262 minutes with a mean life of 606606 minutes. If the claim is true, in a sample of 9999 batteries, what is the probability that the mean battery life would be greater than 619619 minutes? Round your answer to four decimal places.

Answer 1

0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes

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 = 606, \sigma = 62, n = 99, s = (62)/(√(99)) = 6.23`

What is the probability that the mean battery life would be greater than 619 minutes?

This is 1 subtracted by the pvalue of Z when X = 619. So

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

By the Central Limit Theorem

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

`Z = (619 - 606)/(6.23)`

`Z = 2.09`

`Z = 2.09` has a pvalue of 0.9817

1 - 0.9817 = 0.0183

0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes