# Help on Prob Dist Question? Have I done it correctly?

this question is from essential econometrics 3rd edition by Damodar N Gujarati Chp 4 (question 4.12) ------------------------------- Question ------------------------------- The grade point average in an econometrics test was normally distributed with a mean of 75. In a sample of 10 percent of students it was found that the grade point average was greater than 80. Can you tell what the stand deviation of the grade point average was? ------------------------------- My solution ------------------------------- z = (x-u)/sd sd = (x-u)/z as 10% of the population is > 80 so the corresponding z value is 1.285. putting the values in the above sd eq sd = 3.89

come on guys … i will appreciate your help … Joey anyone plz???

I don’t even understand the question. “The grade point average in an econometrics test was normally distributed with a mean of 75.” (80 - 75)/(sigma/sqrt(10))) But you drew the sample, which might be a probability 1/100000000000 event so the numbers don’t help us here.

The key is that you are told that the distribution is normal. That means that the CDF should be equal to 0.9 at a value of 80%, and if you use =normsinv(.9) in excel, that works out to be 1.2816. Thus, mean +/- 1.2816 standard deviations should leave 10% in the right tail. To you, that means 75 + 1.28156(stdev)= 80, so your stdev = 5/1.28156 = 3.9 which is how you computed it.

Except it didn’t say that the highest 10% scored greater than 80. It said a sample of 10% scored greater than 80. Big difference. Edit: should be “averaged greater than 80”

JDV - you are correct! I missed that in my haste.

“Can you tell what the standard deviation was?” No, not with that info. I saw this post and figured that maybe a of the more math specialist types could find an answer. But I don’t see how you would do it. Now, if you took a whole bunch of samples, and you know that 10% of those samples scored above 80, you might get somewhere, if you know the sample size. But that wouldn’t depend on the population being normal.

I am confused with the same logic that you said above but if question would have specified that the top 10% scored >80 it would haven been very easy to solve it. I have taken that assumption. I think that is the assumption you have to take. I can see no other solution. If somebody have the solution manual of the book, can he/she give me the correct answer?

I think the fact that it asks “Can you tell what the stand deviation of the grade point average was?” is indicative of the fact that the answer is “No,” for reasons JDV stated in his post. If it were possible to answer, the question would have asked “What is the standard deviation?” or “Can you tell what the stand deviation of the grade point average was, and if so, what is it?” It’s a conceptual question, not a math question. They just didn’t format it in the CFA Yes/Yes Yes/No No/Yes No/No.

bchadwick your solution makes sense here … It was a tricky one

Hold on now, I have been thinking of this. If you have a normal distribution with a mean of 75, it is quite possible, based on the value of the standard deviation, that you can take a random sample of 10% of the students and have a mean above 80. It should be possible to figure the bounds on a standard deviation from this. Let’s say that I have 100 students, and I use matlab (or excel) to generate random normal variables with mean 75, stdev 5: I just did this and got the following samples (where I have sorted the results in descending order). It’s quite possible to take 10 random samples and get a mean of 80. The mean of the top ten, for example, is 83ish. So I would approach it by finding the minimum standard deviation that would give you a mean of the top 10% = 80. Yes, I know it’s a random sample and you wouldn’t expect to pick the top ten as your sample, but this would be a minimum bound on the standard deviation. For example a standard deviation of 3 would not give you this result, and just iterating a bit says that 3.5 is pretty close. Datapoint Grades 1 86.54643743 Mean 74.60918345 2 85.68154211 Stdev 4.784512826 3 84.31700307 4 83.85050446 5 82.6050662 6 81.39042073 7 81.38226237 8 81.34890924 9 81.18277826 10 81.13723995 11 81.06722247 12 80.36343134 13 80.32059574 14 80.27824194 15 79.95057436 16 79.72099863 17 79.56570409 18 79.44586309 19 78.40219292 20 78.34077517 21 78.17637067 22 78.1017507 23 77.9346928 24 77.75592356 25 77.6231934 26 77.49912834 27 77.42748854 28 77.40067911 29 77.31024006 30 76.98287659 31 76.89611811 32 76.62772992 33 76.58267907 34 76.40440244 35 76.3083123 36 76.30404199 37 76.2154735 38 76.0944956 39 75.86832834 40 75.51712223 41 75.42995297 42 75.35686432 43 75.27970339 44 75.0486708 45 75.03762243 46 74.99591485 47 74.97497463 48 74.94357219 49 74.94106338 50 74.93411664 51 74.80780618 52 74.60839402 53 74.43388005 54 74.41940753 55 74.33432775 56 74.08959066 57 73.77306852 58 73.75281858 59 73.74396313 60 73.71191442 61 73.67993323 62 73.62666507 63 73.6189107 64 73.39497654 65 73.26408405 66 72.99166633 67 72.53456041 68 72.38720349 69 72.29395335 70 72.26091927 71 72.09867999 72 71.99293937 73 71.84360172 74 71.77660542 75 71.518976 76 71.48222869 77 71.47849136 78 71.43957274 79 71.08553478 80 70.96175435 81 70.29313903 82 69.90931392 83 69.89429157 84 69.8551245 85 69.50079773 86 69.46465053 87 69.40480212 88 69.12719859 89 68.84181733 90 68.71704946 91 68.64749898 92 68.33234635 93 67.95235755 94 67.41230434 95 66.68196774 96 66.67994562 97 64.97718339 98 64.39786656 99 63.37394436 100 63.17705076

This is a classic example of people making far too many assumptions about normal distributions, and coming up with methods out of thin air to generate some fact about a data set. The question says that a sample of 10% of the students was taken. What if the population was 10? Then the sample has one value, which is greater than 80, and would also be the mean of that sample. Does the knowledge that one out of 10 tests was greater than 80 tell you what the std. dev. for the population is? NO!!