Question

Twelve percent of all U.S. households are in California. A total of 1.3 percent of all U.S. households earn more than $250,000peryear,whileatotalof3.3percentofallCaliforniahouseholdsearnmorethan$250,000 per year

(a) What proportion of all non-California households earn more than \(250,000 per year?

(b) Given that a randomly chosen U.S. household earns more than \)250,000 per year, what is the probability it is a California household

Short answer

The proportion of all non-California households earning more than $250,000 per year is$=0.01027$

The probability is a California household$=0.3046$

Step-by-step solution

Given information (part a)

The object is to calculate the proportion of all non-California households earning more than $\$250,000$per year using the given information. $12\%$of all U.S. households are in California and $1.3\%$of all U.S. households earn more than $\$250,000$per year. Also,$3.3\%$all California households earn more than$\$250,000$ per year.

Conditional probability expression (Part a)

The proportion of all U.S. households earn more than $\$250,000$per year is

$P(>250K)=\frac{1.3}{100}$

The proportion of households in California is $P\left(C\right)=\frac{12}{100}$

Final answer (Part a)

The proportion of all non-California households earning more than$\$250,000$per a year is $P(>250K\mid \overline{C})$

The proportion of all U.S. households earning more than $\$250,000$per year can be shows using the BAYES's formula as

$P(>250K)=P\left(C\right)P(>250K\mid C)+P\left(\overline{C}\right)P(>250K\mid \overline{C})$

Put the values and simplify to get the proportion of all non-California households earning more than$\$250,000$

$\frac{1.3}{100}=\left[\frac{12}{100}\times \frac{3.3}{100}\right]+\left[\frac{88}{100}\times P(>250K\mid \overline{C})\right]$

$0.013=[0.12\times 0.033]+[0.88\times P(>250K\mid \overline{C}\left)\right]$

$P(>250K\mid \overline{C})=\frac{0.013-0.00396}{0.88}$

$=0.01027$

The proportion of all non-California households earning more than per year is$0.01027$

Given information (part b)

We have to give that a randomly chosen U.S. household earns more than $\$250,000$per year, then the probability it is a California household is$P(C\mid >250K)$

Conditional probability  expression (Part b)

We have the conditional probability, the event$P(C\mid >250K)$can be expressed as

$P(C\mid >250K)=\frac{P(C\cap >250K)}{P(>250K)}$

$=\frac{P\left(C\right)P(>250K\mid C)}{P(>250K)}$

Substitute and simplify values to get the required probability.

Given that a randomly chosen U.S. household earns more than$\$250,000$per year, then the probability it is a California household is

$P(C\mid >250K)=\frac{P\left(C\right)P(>250K\mid C)}{P(>250K)}$

$=\frac{\frac{12}{100}\times \frac{3.3}{100}}{\frac{1.3}{100}}$

$=0.3046$

Final answer (part b)

Finally we get at given randomly chosen U.S. household that earns more than $\$250,000$per year, then the probability it is a California household is $0.3046$