Question

A study of homeowners in the 5th congressional district in Maryland found that their annual household incomes are normally distributed with a mean of $41,182 and a standard deviation of $11,990 (based on data from Nielsen Media Research). If an advertising campaign is to be targeted at those whose household incomes are in the top 20%, find the minimum income level for this target group.

Answer 1

Answer:

The minimum income level for this target group is of $51,253.6.

Step-by-step explanation:

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:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of $41,182 and a standard deviation of $11,990

This means that $\mu = 41182, \sigma = 11990$

Find the minimum income level for this target group.

The 100 - 20 = 80th percentile, which is X when Z has a p-value of 0.8, so X when Z = 0.84.

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

$0.84 = \frac{X - 41182}{11990}$

$X - 41182 = 0.84*11990$

$X = 51253.6$

The minimum income level for this target group is of $51,253.6.