Question

For the total population of a large a southern city, mean family income is

$34,000, with a standard deviation (for the population) of $5,000. Imagine
that your are taking a subsample of 200 city residents. What is the Z score for
$37,000 ?

Answer 1

The Z score for $37,000 is 8.49.

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 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.

Mean family income is $34,000, with a standard deviation (for the population) of $5,000.

This means that
`\mu = 34000, \sigma = 5000`

Subsample of 200 city residents.

This means that
`n = 200, s = (5000)/(√(200)) = 353.55`

What is the Z score for $37,000 ?

This is Z when X = 37000. So

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

By the Central Limit Theorem

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

`Z = (37000 - 34000)/(353.55)`

`Z = 8.49`

The Z score for $37,000 is 8.49.