Question

Suppose that a population is known to be normally distributed with £ = 2400 and € = 210. Of a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,500.

Answer 1

0.0885 = 8.85% probability that the sample mean will exceed 2,500.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

In this question:

`\mu = 2400, \sigma = 210, n = 8, s = (210)/(√(8)) = 74.25`

Calculate the probability that the sample mean will exceed 2,500.

This is 1 subtracted by the pvalue of Z when X = 2500. So

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

By the Central Limit Theorem:

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

`Z = (2500 - 2400)/(74.25)`

`Z = 1.35`

`Z = 1.35` has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the sample mean will exceed 2,500.