You are checking out an overnight camp for your kids for next year. On your family visit to the camp you ask 30 randomly encountered kids what is their bunk size and record all the answers. Surprisingly, the answers deviate quite a lot. How can you figure out (approximate) average bunk size from all these numbers?

Image by New York YMCA camp, distributed under CCL.

This is a tricky puzzle and therefore I am giving one whole week for contemplation and answer submission. At the present moment I don't know either how this should be solved but I do know how this should not be solved. So, let's figure it out together. Answer ideas accepted any time until next Friday July 22nd, on our Family Puzzle Marathon. They will be hidden till then and everyone who suggested something reasonable will get a puzzle point.

## 9 comments:

Add up the measurements and divide by 30.

That is, if you believe the children, and if you assume that their measurements are correct. If however you cannot rely on the children's accuracy, then you'd better measure the bunks yourself, length and width in inches probably, add the lengths and divide by the number of bunks measured for average length; add your widths together, do similar division for average width. And then multiply L x W.

30 randomly selected children, or 30 randomly selected bunks (cots) might not be a large enough sample, depending on the size of the camp.

Further, there may indeed be a variety of cot sizes for the various age groups that can be at a summer camp.

Work out the mean, variance & stadard deviation. Choose a confidence level from a t table work out the upper and lower limits for the bunk size. If your're worried about whether your kid will fit comfortably then the lower limit is the one your worried about.

I will state the central limit theorem.

Whenever a random sample of size n Is taken from a distribution with mean "u"

And variance"", then the sample mean will be approximately normally distributed with mean "u"and variance ""/n.

(I do not have the symbol on the ipad for the variance)

The bigger the n , the better it is.

We know that 95%would be in two standard Deviation from the mean so I will figure out a size of bed that fit all this group.

Then, I would take a couple of bed above average and below average( 2.5%).

Oh, I am so sorry...I thought kid size to find the average size of a bunk bed. I may try later..

Ok

We take into consideration the bunk size and not the kid size.

I would consider the mode of these 30 numbers.

I would go for simplicity. Given a lot of deviation, let's assume that we are dealing with a symmetrical bell curve. 30 answers is a pretty good statistical sample to represent most of the curve, so I think all that's needed is a quick scan through the list to find the largest and the smallest and to compute the average of this pair.

I have only iPhone connection today - puzzle solution and the newletter will be coming later on Friday.

This seems to be a very applicable yet tricky puzzle.

To agree that we are talking about the same “average” let’s consider two cases:

Case 1: there are two bunks, sizes 2 and 8 kids. Bunk size average is (2+8)/2=10/2=5

Case 2: there are three bunks, sizes 2, 8, 8 kids. Bunk size average is (2+8+8)/3=18/3=6

First, let’s see why a simple and the most intuitive answer – average of all the kids’ responses is not a solution. Our random sample will have more kids from the larger bunks shifting the average to the right.

If we take Case 1 and assume that we ask every single kid what is his/her bunk size, the answers will be: 2 (twice) and 8 (8 times). Simple average will be: (2x2+8x8)/10=68/10=6.8 And we know the correct bunk size average in the Case 1 is 5.

Another simple answer – take only one answer from each bunk size and average them all – is unfortunately also not a solution. It will work well for Case 1, however for Case 2 with two bunks of size 8 we will get (2+8)/2=5 while we know that the correct solution is 6.

OK, we know what is NOT a solution. What is THE solution then?

Note that our random kids sample gives us chances of someone to be in a bunk of size N.

For Case 1, if we ask 10 kids and receive 2 answers of size 2 and 8 answers of size 8, chances of a kid to be in a bunk of size 2 is 2/10 and of size 8 is 8/10.

For Case 2, if we ask 9 kids (out of total 18 kids) and proportionally receive 1 answer of size 2 and 8 answers of size 8, chances of a kid to be in a bunk of size 2 is 1/9 and of size 8 is 8/9.

In general case:

Chances of a kid to be in bunk of size N = # of random kids from bunk of size N / total # of randomly asked kids.

Denote the above with (*) as it will come handy later.

To save space let’s agree that we define as SUM(i) as Sum_over_all_bunk_sizes_of_i.

Average bunk size (call it ABS) =

SUM (#bunks_of_this_size X bunk_size) /

SUM ( #bunks_of_this_size)

#bunks_of_each_size = chances_of_a_kid_to_be_in_bunk_of_this_size X total_num_kids / bunk_size

Substituting (*) here for chances… we get:

#bunks_of_this_size = ((# of random kids from bunks of size N / total # of randomly asked kids) X total_#_kids) / bunk_size

Denote: total_#_kids/ total # of randomly asked kids = A

#bunks_of_this_size = A X # of random kids from bunks of size N / bunk_size

Then our ABS =

SUM ( A X # of random kids from bunks of this size) /

SUM(A X # of random kids from bunks of this size / bunk_size )

Taking A away:

ABS = SUM(# of random kids from bunks of this size) /

SUM( # of random kids from bunks of size N / bunk_size )

Our nominator SUM is actually just the total number of random kids asked.

Therefore:

ABS = total # of random kids asked / SUM (# of random kids from bunks of size N / bunk_size )

For Case 1 we will have: 10/ (2/2+8/8) = 10/2 =5 Correct!

For Case 2 we will have: 9/ (1/2+8/8) = 9/(3/2)=18/3=6 Correct!

Maria

Needless to say that everyone who dared to submit their thoughts to this puzzle receives a puzzle point - Tom, anne-marie and Ilya. Anonymous - you will get one as well if you will reveal your name.

No new puzzle today - enjoy the summer sunshine and vacations till September.

Maria

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