Question

The Federal Reserve System publishes data on family income based on its Survey of Consumer Finances. When the head of the household has a college degree, the mean before-tax family income is $ 85,050. Suppose that 56% of the before-tax family incomes when the head of the household has a college degree are between $75,000 and $95,100 and that these incomes are normally distributed. What is the standard deviation of before-tax family incomes when the head of the household has a college degree

Answer 1

The standard deviation is $13,052.

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 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.

When the head of the household has a college degree, the mean before-tax family income is $ 85,050.

This means that
`\mu = 85,050`

Suppose that 56% of the before-tax family incomes when the head of the household has a college degree are between $75,000 and $95,100 and that these incomes are normally distributed.

They are equally as far from the mean, one above, and one below. This means that when
`X = 95100`, Z has a pvalue of 0.5 + (0.56/2) = 0.78. So when X = 95100, Z = 0.77. We use this to find
`\sigma`.

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

`0.77 = (95100 - 85050)/(\sigma)`

`0.77\sigma = 10050`

`\sigma = (10050)/(0.77)`

`\sigma = 13052`

The standard deviation is $13,052.