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AP Stats Question

Registered User Posts: 464 Member
edited January 2008
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!
Post edited by Eravial08 on

Replies to: AP Stats Question

• Registered User Posts: 308 Junior Member
Using binomial cumulative distribution, I got D)0.5248, because it is at least 2 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) = 0.5248

I 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.
• Registered User Posts: 99 Junior Member
e) The answer cannot be computed from the information given.

The 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.
• Registered User Posts: 4,535 Senior Member
lol....so who's right? I also thought that the answer was 'e'...but i wasn't too sure.
• Registered User Posts: 842 Member
tw14 is right, but I think the question may be a typo?
• Registered User Posts: 464 Member
We had this problem on a test in class, and it is supposedly from a past AP Test. I think the answer is 'e' and I have checked with the other top students (750+ Math SATI and MathIIC) in my stats class, and all of them agree. However, our teacher asserts that 'd' is the correct answer, and some students defend that answer. Consequentially, everyone who answered 'e' was marked wrong and docked points.

You 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?
This discussion has been closed.