ok24by7 wrote:How to solve this attached problem?

Thanks.

i dont see any attachment . did u forget to add the attachment?

Think of the median as the middle number not as the average. The question is phrased difficultly in that it seems like it is telling you the median number is 73.

For example the median of this set of numbers:
1,2,3,4,4,4,4 is 4 b/c it is in the middle of the set.
However, if the set was:
1,2,3,4,4,4 is 3.5 b/c when you have an even number in the data set, you average the middle 2 numbers to get the median.

The problem you posted is a little tricky since it seems like the median would be closer to the average, but it is the middle number instead. I like to cross a numbers off of each end until if find the middle, and by doing that you see that the majority of the numbers are concentrated to the higher end.

Let me know if you need any more help on that.

There are 73 individual scores in total. Call them -- S1, S2,S3........S73.

Rewrite the score table as follows:
50 - 59 : S1, S2
60 - 69 : S3, S4,..,S12
70 - 79 : S13,S14,...S28
80 - 89 : S29, S30,...S37,...S55

The median score is S37 (37th score is the mid point of 73 scores)

Looking at the table above, 37th score falls in the 80 - 89 interval

This how i have understood it. A tricky question indeed!

Fundamentally, this question will be solved by calculating cumulative frequency.

I lay my cards on same explanation as given below..

sg4931 wrote:There are 73 individual scores in total. Call them -- S1, S2,S3........S73.

Rewrite the score table as follows:
50 - 59 : S1, S2
60 - 69 : S3, S4,..,S12
70 - 79 : S13,S14,...S28
80 - 89 : S29, S30,...S37,...S55

The median score is S37 (37th score is the mid point of 73 scores)

Looking at the table above, 37th score falls in the 80 - 89 interval

This how i have understood it. A tricky question indeed!

some times ( like this one) i believe that it´s better to number all the scores like:

1 for 50-59
2 for 60-69
3 for 70-79
4 for 80-89
5 for 90-99

of course in the heat of the moment this will be something like (1,2,3,4,5)

112222222222333333333333333444444444444444444444444444444444444444444444555555555555555555555555


count 37 digits and you will find the right interval !

Easiest solution. How many scores? 73. The median will be the 37th value. (36 below and 36 above). The 37th value is in the 4th score interval, 80-89.

How I solved it:

Add up total number of scores and divide by 2 (73/2 = 36.5)

Which one has the 36.5th number in it?

50-59 has numbers 1-2
60-69 has numbers 3-12
70-79 has numbers 13-28
80-89 has numbers 29-55
90-99 has numbers 56-73

29 < 36.5 < 55