Sign Up For Free

**Join for FREE**,
and start talking with other members, weighing in on community polls,
and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one!)

- Reply to threads, and start your own
- Create reports of your
**campus visits** - Share college
**photos**and**videos** **Find your dream college**, save your search and share with friends- Receive our
**monthly newsletter**

Home
/
College Discussion / College Admissions and Search / SAT and ACT Tests & Test Preparation / AP Tests Preparation

Eravial08
Posts: **464**Registered User Member

There is a discrepancy in my AP Statistics class as to the correct answer to this question. What is your answer, and why?

(From the binom/geomet, pdf/cdf chapter: )

In a certain large population, 40% of households have a total annual income of $70,000. A simple random sample of 4 of these households is selected. What is the probability that 2 or more of the households in the survey have an annual income over $70,000?

a) 0.3456

b) 0.4000

c) 0.5000

d) 0.5248

e) The answer cannot be computed from the information given.

Thanks!

(From the binom/geomet, pdf/cdf chapter: )

In a certain large population, 40% of households have a total annual income of $70,000. A simple random sample of 4 of these households is selected. What is the probability that 2 or more of the households in the survey have an annual income over $70,000?

a) 0.3456

b) 0.4000

c) 0.5000

d) 0.5248

e) The answer cannot be computed from the information given.

Thanks!

Post edited by Eravial08 on

## Replies to: AP Stats Question

308Registered User Junior Memberat least2 of the four, so you would do (I'm using TI notation) 4nCr2(0.4^2)(0.6^2) + 4nCr3(0.4^3)(0.6) + 4nCr4(0.4^4) = 0.5248I used binomial distribution because the probability is one of two options: greater than $70k or less than $70k. It's cumulative because it's 2 houses, 3 houses, or 4 houses.

99Registered User Junior MemberThe problem states "40% of households have a total annual income of $70,000." It does not state that 40% have an income of $70,000 or less, or $70,000 or greater. It gives one fact: 40% have an income of $70,000. It does not tell us what percentage have an income of $40,000, or $50,000, or $60,000, or $80,000, etc., etc.

What if it read "40% of households have 3 members?" That does not tell us anything about what percentage of the households have 1 member, 2 members, 4 members, 5 members, etc.

What if it read "40% of houses are painted green?" Does that mean that 40% of houses have a color with a wavelength of 570 nm or less, which would include blue and violet houses? Of course not.

Answer e) is often a choice on AP problems because it is important for students to base their response on the facts presented, not on assumptions.

By the way, if the problem had stated that the median income was $70,000, then there would have been sufficient information presented to solve the problem.

4,535Registered User Senior Member842Registered User Member464Registered User MemberYou get d by assuming that the probability of one household having an annual income under $70,000 is .40. Entered into a TI-83, this would be binomcdf(4, .40, 1).

However, am I correct in that you cannot assume that "have a total income of $70,000" actually means "have a total income of $70,000 or less"? I thought that since you were only given the one value - the probability for ONLY $70,000 - it was impossible to determine the distribution other than at the $70,000 mark.

Can anyone explain why answer 'd' would be correct?