Question

If a normally distributed population has a mean (mu) that equals 100 with a standard deviation (sigma) of 18, what will be the computed z-score with a sample mean (x-bar) of 106 from a sample size of 9?

Answer 1

Z = 1

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
`\mu` and standard deviation
`\sigma`, the z-score 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 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
`\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.

Mean (mu) that equals 100 with a standard deviation (sigma) of 18

`\mu = 100, \sigma = 18`

Sample of 9:

This means that
`n = 9, s = (18)/(√(9)) = 6`

What will be the computed z-score with a sample mean (x-bar) of 106?

This is Z when X = 106. So

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

By the Central Limit Theorem

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

`Z = (106 - 100)/(6)`

`Z = 1`

So Z = 1 is the answer.