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Re: IR question from Princeton Review 2013.. help needed [#permalink]

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09 Jun 2012, 23:03

5

This post received KUDOS

Here are my notes. How did I do? Helpful?

Section 1:

A: No. Only implies that it's happened before, possibly last year.

B: Yes. Midwest Mean = West Mean = min(all means).

C: No. Could be Level 2 + anybody else, probably also ROMs.

D: Yes. He/she can delegate.

Section 2:

A: We know that BB is 50% of difference between ticket price and that region's mean. For West, mean is 200. Only tickets purchases below 1 SD (25) are the 18 purchased at 150. A few different ways to do the math here, I'll do this: 200 - 150 = 50 50 * 50% = 25 25 * 18 = (25*20) - (25*2) = 500 - 50 = 450 :. West will receive BB of $450. :. True

B: Mid-Atlantic info from chart: 500 sample size, 350 mean, and SD of 50. I'm not convinced about this one because I'm not sure whether this is a normal distribution. In a SD, 68% of the tickets would be within one standard deviation and 95% of the tickets would be within two. Let's run with that. So $450 is two standard deviations away, so is 5% / 2 of the sample size greater than 20? It's important that we also divide 5% by 2 because that's just the portion that is on the upper end of the normal distribution. 500 * 0.1 = 50 * 0.1 = 5 :. 5 = 1% of 500 :. 5% of 500 = 25 :. 2.5% = 12.5 :. 12.5 < 20 :. False

C: Same process as in B. 400 sample size, 300 mean, and 50 SD. This time we're only talking about one SD though. Is 32% / 2 of the sample size greater than 50? Let's estimate because it's timed. 400 * .1 = 40 * 3 = 120 = 30% of 400 :. 15% of 400 = 60 :. 60>50 :. True

Re: IR question from Princeton Review 2013.. help needed [#permalink]

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14 Jun 2012, 20:27

It is worth noting that statistics have been defined as "normally distributed" That's why , 68% will fall under 1 Deviation (32% lie outside) 95% will fall under 2 Deviation 99.5% will fall under 3 Deviation 99.999997% will fall under 6 deviation Had it not been the "normal distribution", answer of C (II question) might have been different.
_________________

Re: IR question from Princeton Review 2013.. help needed [#permalink]

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30 Dec 2015, 08:39

Hello,

Can someone help me explain how the precentage of the normal deviation is determined? I am not familiar on how you receive the % of 2 and 16 % for the sections B and C under Section 2.

Re: IR question from Princeton Review 2013.. help needed [#permalink]

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27 Feb 2016, 21:23

mdellaratta wrote:

Hello,

Can someone help me explain how the precentage of the normal deviation is determined? I am not familiar on how you receive the % of 2 and 16 % for the sections B and C under Section 2.

Still too new to post the link, so unable to share the pic here. Anyway, just google "34 14 2 standard deviation" and the answer is in GRE SparkNotes. I also learned this new piece info from the same question.

gmatclubot

Re: IR question from Princeton Review 2013.. help needed
[#permalink]
27 Feb 2016, 21:23