Question

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?

Answer 1

a) P(X < 14) = 0.0228.

b) 49 batteries must be tested.

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 = 15, \sigma = 3, n = 36, s = (3)/(√(36)) = 0.5`

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

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

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

By the Central Limit Theorem

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

`Z = (14 - 15)/(0.5)`

`Z = -2`

`Z = -2` 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.

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

`-2.327 = (14 - 15)/(s)`

`-2.327s = -1`

`2.327s = 1`

`s = (1)/(2.327)`

`s = 0.43`

We know that

`s = (\sigma)/(√(n))`

So

`0.43 = (3)/(√(n))`

`0.43√(n) = 3`

`√(n) = (3)/(0.43)`

`(√(n))^(2) = ((3)/(0.43))^(2)`

`n = 48.7`

Rounding up

49 batteries must be tested.