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If you have 9 golfers with mean 12 with 0 as lowest and 24 as highest, it is impossible to get a stdev as low as 3 -- even if all of the other golfers were at the mean. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

If you have 9 golfers with mean 12 with 0 as lowest and 24 as highest, it is impossible to get a stdev as low as 3 -- even if all of the other golfers were at the mean.

Thanks! You were absolutely right. I made a revision as a result. Kudos Akamai!

A "Two some" is to be chosen out of an eligible pool of 9 golfers to play at a charity tournament. The best golfer in the pool has a handicap of 0 while the worst golfer has a handicap of 24. No two golfers have the same handicap. The median and mean handicap are 12 and each of the golfer's handicap from worst to best can be derived from the sequence Tn=Tn-1 + 3. What is the probability that the median handicap of the golfers chosen from the pool will have a median handicap less than 15?

as far as I understand the question you have normal distribution with mean and median 12, SD 4, since 15 < 1 SD from mean and you want the median score to be less than 15, you have 50%+34%=84% probability

Forget the Standard Deviation - i took that out. You should be able to figure out all the handicaps based on the sequence formula. I'll let you take it from there!

although i can not understand your exp prob very well think that this is what you mean: the median means that it is in the middle between any 2 golfers selected, so 2 of 9 in 36 ways, if select 24( we can not select 21,18,15,12,9 cause the median is>15) -3 ways, 23-3ways, 18-4 ways, 3 out of 6 (0,3,6,9,12,)=20 ways or total 30 ways, or30/36= 5/6 or 83%..hope makes sense