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Question:

An analysis of economic data shows that the annual income of a randomly chosen person from country A has a mean of 18,000 and a standard deviation of 6,000, and the annual income of a randomly chosen person from country B has a mean of 31,000 and a standard deviation of 8,000.

One hundred individuals are chosen at random from Country A and 100 from Country B.

Find the approximate probability that the average annual income from the group chosen from country B is at least 15,000 larger than the average annual income from the group chosen from country A.

Multiple choice options are: (A)0.101 (B)0.023 (C)0.370 (D)0.659 (E)0.805

My Attempt:

I let B represent the avg annual income group B which would be 31,000 and A the avg annual income from group A which is 18,000.

Then I need to find P(B >= 15,000 + A). Using CLT I got P(Z >= (18,000+15,000-31,000)/8,000) which is P(Z >= .25) which is .4013.

Not sure what I'm doing wrong here

user118342
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1 Answers1

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You want to divide by 10,000 and not $8,000$. Best way to see this is as follows:

Let $$X = B-A$$ Then $$ E[X] = 31,000-18,000 = 13,000 $$ $$ \text{Var}[X] = \text{Var}[B] + \text{Var}[A] = 6000^2 +8000^2 =10000^2 $$ So $$ Z = \frac{15000-13000}{10000} = 0.2 $$ You can complete the rest.

user44197
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    I don't see any of your choices as being valid. Did you write the question correctly? – user44197 Dec 31 '13 at 21:50
  • Yeah I checked and the question/MC options are correct. So you standardized the difference between A and B? – user118342 Dec 31 '13 at 22:02
  • Yes. The hypothesis is about $B-A$ just stated differently in English. One way to see this in similar problems, move all random variables to one side and all constants, non-random variables to the other side of an inequality. Then define $X$ to be the random part – user44197 Dec 31 '13 at 22:04
  • I think this would be the correct answer if a single person were chosen from each population: but you are missing the bit about "One hundred individuals are chosen at random..." – leonbloy Dec 31 '13 at 22:25
  • I think I found a small mistake with the standardization of B-A. If B represents the annual average income of a group of 100 randomly selected individuals then B should follow a normal distribution with mean 31,000 and standard deviation 8000/(sqrt(100)) which is 800. A would have SD 600. Then X would have SD 1000 and so Z = 2. Final answer would then be .0228? – user118342 Jan 04 '14 at 18:15