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09 Jun 2012, 21:50
Kudos
Bookmarks
Here is another IR question from PR 2013.
Solved most of the questions except some from the 'True-False' questions.
Can someone please help.
Thanks in advance
Solved most of the questions except some from the 'True-False' questions.
Can someone please help.
Thanks in advance
09 Jun 2012, 23:03
Kudos
Bookmarks
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
D:
Midwest: 300 sample, 200 mean, 25 SD
:. Mean - 2*SD = 200-50 = 150
Plains: 400 sample, 300 mean, 75 SD
:. Mean - 2*SD = 300-150 = 150
:. False
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
D:
Midwest: 300 sample, 200 mean, 25 SD
:. Mean - 2*SD = 200-50 = 150
Plains: 400 sample, 300 mean, 75 SD
:. Mean - 2*SD = 300-150 = 150
:. False
14 Jun 2012, 13:50
Kudos
Bookmarks
thanks a lot !! It helps me a lot !!
