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

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


J
Asked by 1 year ago
240 points

Twelve percent of all U.S. households are in California. A total of 1.3 percent of all U.S. households earn more than \$250,000 per year, while a total of 3.3 percent of all California households earn more than \$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?

probability
bayes theorem

2 Answers

Answered by 1 year ago
8.7k points

Let,

C the event that a household is in California

M the event that a household in the U.S. makes more than $250,000

What proportion of all non-California households earn more than $250,000 per year?

We have that,

$$P(C \cap M)=P(C)P(M)=0.120.033=0.00396 = 0.396\%$$

households live both in California and make $250,000 per year

following this logic we have $P(M \cap C^c) = 1.3\% - 0.396\% = 0.904\%$

non-Californians who make over $250,000

and proportionally to the population of non-Californians,

$$\frac{0.904\%}{88\%}=0.0103$$

or 1.03% of non-California households earn more than $250,000 per year

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

using Bayes' theorem,

$$P(C|M)=\frac{P(M|C)P(C)}{P(M|C)P(C)+P(M|C^c)P(C^c)}$$

$$\longrightarrow \frac{0.033(0.12)}{0.033(0.12)+0.00904}=0.3046$$

where I have used the fact that $P(M|C^c)P(C^c) = P(M \cap C^c)=0.904%=0.00904$

Z
Answered by 8 months ago
0 points

Oh Snap! This Answer is Locked

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

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