A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 2916 with a mean life of 518 minutes. If the claim is true, in a sample of 81 batteries, what is the probability that the mean battery life would be greater than 526.4 minutes

0.0808 = 8.08% probability that the mean battery life would be greater than 526.4 minutes

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

The design engineer claims they have a variance of 2916 with a mean life of 518 minutes.

This means that

Sample of 81 batteries

This means that

What is the probability that the mean battery life would be greater than 526.4 minutes?

1 subtracted by the p-value of Z when X = 526.4. So

By the Central Limit Theorem

has a p-value of 0.9192

1 - 0.9192 = 0.0808

0.0808 = 8.08% probability that the mean battery life would be greater than 526.4 minutes

98%

step-by-step explanation: