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The Federal Reserve System publishes data on family income based on its Survey of Consumer Finances....

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 $ 84,950. Suppose that 58% of the before-tax family incomes when the head of the household has a college degree are between $74,800 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? (Round the value of z to 2 decimal places. Round your answer to 2 decimal places.)

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Answer #1

\\ $ Mean $ (\mu) = \$84,950 \\\\ $ Let standard deviation be $ \sigma \\\\ $ 58\% are between \$74,800 and \$95,100 $ \\\\ \Rightarrow P(\ 74,800 < x < 95,100\ ) = 0.58 \\\\ $ As $ z = \dfrac{ x - \mu}{\sigma} \\\\ \Rightarrow P(\ \dfrac{74,800 - 84,950}{\sigma} < z < \dfrac{95,100 - 84,950}{\sigma}\ ) = 0.58 \\\\ \Rightarrow P(\ -\dfrac{10,150}{\sigma} < z < \dfrac{10,150}{\sigma}\ ) = 0.79 - 0.21 \\\\ \Rightarrow (\ -\dfrac{10,150}{\sigma} <z < \dfrac{10,150}{\sigma}\ )= P(\ z < 0.81) - P(\ z < -0.81) \\\\ ( P( z < 0.81) $ and $P( z < -0.81) $ are closest to 0.79 and 0.21 $ \\ $ respectively from Z table $) \\\\ \Rightarrow (\ -\dfrac{10,150}{\sigma} <z < \dfrac{10,150}{\sigma}\ )= P(\ -0.81 < z < 0.81 \ ) \\\\ \Rightarrow \dfrac{10,150}{\sigma} = 0.81 \hspace{15 mm} ( $ Comparing both sides $ ) \\\\ \Rightarrow \sigma = \dfrac{10,150}{0.81} \approx \textbf{\$12,530.86}

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