Question

The average amount of water in randomly selected 16-ounce bottles of water is 16.15 ounces with a standard deviation of 0.45 ounces. If a random sample of thirty-five 16-ounce bottles of water are selected, what is the probability that the mean of this sample is less than 15.99 ounces of water? Answer: (round to 4 decimal places)

Answer 1

0.0177

Explanation:

New SD = .45/SQRT(35) = 0.076064

P(x < 15.99), in Excel

=NORM.DIST(15.99,16.15,0.076064,TRUE)

Answer 2

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

Explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 question, we have that:

`\mu = 16.15, \sigma = 0.45, n = 35, s = (0.45)/(√(35)) = 0.0761`

What is the probability that the mean of this sample is less than 15.99 ounces of water?

This is the pvalue of Z when X = 15.99. So

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

By the Central Limit Theorem

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

`Z = (15.99 - 16.15)/(0.0761)`

`Z = -2.1`

`Z = -2.1` has a pvalue of 0.0179

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.