Stats + probability Amazing question!

[magnus opus]

Five cards are to be selected at random from 10 cards numbered from 1 to 10. How many ways are possible that the average of the five numbers selected will be greater than the median?

have no clue how to solve this one under two minutes. please help.

[shovan85]

5 numbers can be selected from 10 in C(10,5) = 252
Median is 5.5
we need mean> 5.5 so sum > 27.5
So sum of 5 numbers >= 28 as 27.5 is not achievable all integers.

When 1234 comes in first 4 pick last pick does not matter so 6 ways (5,6,7,8,9,10)
When 1345 comes in first 4 pick last pick does not matter so 5 ways (6,7,8,9,10:12345 is taken care of previously)
When 1456 comes in first 4 pick last pick does not matter so 4 ways (7,8,9,10)
When 1567 comes in first 4 pick last pick does not matter so 1 ways (8 only 9 and 10 will be > 27.5)

When 2345 comes in first 4 pick last pick does not matter so 5 ways (6,7,8,9,10)
When 2456 comes in first 4 pick last pick does not matter so 4 ways (7,8,9,10)

When 3456 comes in first 4 pick last pick does not matter so 3 ways (7,8,9)

Total 28 ways when the sum will not be > 27.5

Thus Ans will be (252-28)/252 = 224/252
whats the answer?

[GMATGuruNY]

Five cards are to be selected at random from 10 cards numbered from 1 to 10. How many ways are possible that the average of the five numbers selected will be greater than the median?

have no clue how to solve this one under two minutes. please help.

The number of ways to choose 5 cards from 10 choices = 10C5 = 252.

We need to subtract from 252 the number of combinations in which the average = the median:

Median and average of 3:
{1,2,3,4,5}

Median and average of 4:
{1,2,4,5,8} {1,2,4,6,7} {1,3,4,5,7} {2,3,4,5,6}

Median and average of 5:
{1,2,5,7,10} {1,2,5,8,9} {1,3,5,6,10} {1,3,5,7,9} {1,4,5,6,9} {1,4,5,7,8} {2,3,5,6,9} {2,3,5,7,8} {2,4,5,6,8} {3,4,5,6,7}

Median and average of 6:
{1,4,6,9,10} {1,5,6,8,10} {2,3,6,9,10} {2,4,6,8,10} {2,5,6,7,10} {2,5,6,8,9} {3,4,6,7,10} {3,4,6,8,9} {3,5,6,7,9} {4,5,6,7,8}

Median and average of 7:
{3,6,7,9,10} {4,5,7,9,10} {4,6,7,8,10} {5,6,7,8,9}

Median and average of 8:
{6,7,8,9,10}

Total combinations in which the average = the median: 1+4+10+10+4+1 = 30.

Thus, we have 252-30 = 222 combinations in which the average ≠ median.

In half of these combinations, the average will be less than the median; in the other half, the average will be greater than the median.

Thus, the number of combinations in which the average is greater than the median: 222/2 = 111.

Please note that this question would NOT be asked on the GMAT. A bit too time-consuming, I think.

[magnus opus]

yes 111 is the correct answer, but it cannot be done in under two minutes this way.

You cannot and do not have the time to list 30 cases on the D day, and if you miss even one case then by symmetry you would end up doubling the error. ( that is if your realize the symmetry in mean=median cases).

I was thinking of of using the bell curve on such a question if the answer choices are spread out ( far apart to allow approx)

now its easy to figure out that the first case has only one possibility 12345 with mean equal to median.
so 1 case =2%, hence 14%= 7cases, 34%= 17
total cases = sum of all cases on either side of the curve
total cases = 2(1+7+17)= 2 X 25= 50

10 C5= 252

252- 50=202( approx cases for mean> or < median)

only for mean> median =202/2=101 which should be close to 111.


Could you verify this approach...? is it correct to think of the question in these terms.

I have two issues using this approach.
- i knew the answer using the long approach so I was aware of the symmetry in the cases of mean= median
1+4+10+10+4+1 so it is easy to see the peak of the bell curve. However, can we generalise this?

- More important question, can you point out why we do not see an equal number of cases in all three segements namely :
mean=median; mean>median; mean<median

[Ian Stewart]

yes 111 is the correct answer, but it cannot be done in under two minutes this way.

It's clearly not a real GMAT question since there is no reasonable way to do the problem quickly. It isn't worth worrying about.

I was thinking of of using the bell curve on such a question if the answer choices are spread out ( far apart to allow approx)

now its easy to figure out that the first case has only one possibility 12345 with mean equal to median.
so 1 case =2%, hence 14%= 7cases, 34%= 17

If you find yourself tempted to use some properties of normal distributions (bell curves) on the GMAT, stop immediately and find some other approach. Normal distributions are never tested on the GMAT, and if you don't have a very solid grounding in statistics, you'll probably be misapplying what you do know. Here, for example, there's simply no reason to think the numbers are distributed along a bell curve in the first place (they are not, in fact, even 'approximately normally distributed' in the technical sense of that phrase), and I don't see how you've decided that the first case, in which the set is {1,2,3,4,5}, represents 2% of all cases - when you do that, you're simply guessing in advance that there are 50 cases, and that's exactly what we're trying to figure out.

- More important question, can you point out why we do not see an equal number of cases in all three segements namely :
mean=median; mean>median; mean<median

I suppose you could ask the reverse question: why *should* you have an equal number of each case? If I ask a simpler question:

If you pick a random number x from the set {1,2,3,4,5,6,7}, what is the probability the number is greater than 4?

you certainly wouldn't expect it to be equally likely that x < 4, that x = 4 and that x > 4.

If you invent a set of numbers completely at random, it will be very rare that the mean and median are equal, so in the question in the original post, you'd expect the case "median = mean" to be less common that the other possibilities.

[magnus opus]

yes 111 is the correct answer, but it cannot be done in under two minutes this way.

It's clearly not a real GMAT question since there is no reasonable way to do the problem quickly. It isn't worth worrying about.

I was thinking of of using the bell curve on such a question if the answer choices are spread out ( far apart to allow approx)

now its easy to figure out that the first case has only one possibility 12345 with mean equal to median.
so 1 case =2%, hence 14%= 7cases, 34%= 17

If you find yourself tempted to use some properties of normal distributions (bell curves) on the GMAT, stop immediately and find some other approach. Normal distributions are never tested on the GMAT, and if you don't have a very solid grounding in statistics, you'll probably be misapplying what you do know. Here, for example, there's simply no reason to think the numbers are distributed along a bell curve in the first place (they are not, in fact, even 'approximately normally distributed' in the technical sense of that phrase), and I don't see how you've decided that the first case, in which the set is {1,2,3,4,5}, represents 2% of all cases - when you do that, you're simply guessing in advance that there are 50 cases, and that's exactly what we're trying to figure out.

- More important question, can you point out why we do not see an equal number of cases in all three segements namely :
mean=median; mean>median; mean<median

I suppose you could ask the reverse question: why *should* you have an equal number of each case? If I ask a simpler question:

If you pick a random number x from the set {1,2,3,4,5,6,7}, what is the probability the number is greater than 4?

you certainly wouldn't expect it to be equally likely that x < 4, that x = 4 and that x > 4.

If you invent a set of numbers completely at random, it will be very rare that the mean and median are equal, so in the question in the original post, you'd expect the case "median = mean" to be less common that the other possibilities.

well the reason I applied the bell curve was that (as I already mentioned I knew the cases are 30)
there was a pattern to them, 1+4+10+10+4+1. (more at the centre and less at the ends-like a bell)

so the real question was, would cases generally form such a pattern, if so then how can we generalise this?