Question

Jean works for the government and was conducting a survey to determine the income levels of a number of different neighborhoods in a metropolitan area. Based on national data, Jean knows that the mean income level in the country is $40,000, with a standard deviation of $2,000. Jean selected three neighborhoods and determined the average income level. What is the probability that the average income level in the neighborhoods was less than $38,000

Answer 1

Answer:

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

Step-by-step 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 zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Jean knows that the mean income level in the country is $40,000, with a standard deviation of $2,000.

This means that $\mu = 40000, \sigma = 2000$

Jean selected three neighborhoods and determined the average income level.

This means that $n = 3, s = \frac{2000}{\sqrt{3}} = 1154.7$

What is the probability that the average income level in the neighborhoods was less than $38,000

This is the pvalue of Z when X = 38000. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{38000 - 40000}{1154.7}$

$Z = -1.73$

$Z = -1.73$ has a pvalue of 0.0418

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.