A computer battery manufacturer claims their batteries will hold a charge for an average of 15 hours, with standard deviation 3 hours. A random sample of 36 batteries is drawn. Each of the batteries is tested to see how long it holds a charge. Let X = sample mean of the 36 measured times. (a) If the claim is true, what is P(X < 14)? (b) How many batteries must be tested so that the P(X < 14) = 0.01?

a) P(X < 14) = 0.0228.

b) 49 batteries must be tested.

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 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this problem, we have that:

(a) If the claim is true, what is P(X < 14)?

This is the pvalue of Z when X = 14. So

By the Central Limit Theorem

has a pvalue of 0.0228

So

P(X < 14) = 0.0228.

(b) How many batteries must be tested so that the P(X < 14) = 0.01?

Now we want to find the size of the sample for which X = 14 is in the first percentile, that is, Z when X = 14 has a pvalue of 0.01. So it is Z = -2.327.

First we have to find the standard deviation of the sample in this case.

We know that

So

Rounding up

49 batteries must be tested.