Hi,
I had to open the image in another tab. I guess it would be better if you can upload it again.
Coming to the problem -
Let's draw the frequency distribution chart:
It should be arranged in increasing order of values and not frequencies.
Range----Frequency---Cumulative frequency
--0-9--------1--------1
10-19--------4--------5
20-29--------6--------11
30-39--------7--------18
40-49--------2--------20
Median will be average of 10th and 11th terms, both of which are in the range 20-29.
a)We need least possible number of seniors(items) with six pages from median. So, we place other items farther from median.
As the difference between 10th element and 11th element can be maximum 9, these two elements cannot be more than 5 pages within the average of them. So, in any case 10th and 11th elements have to be within 6 pages from median. There are many scenarios for this.I will be posting one case:
Let's consider the case:20,20,20,20,26,27. Median = 26.5 Only 26,27 are within 6 pages from median i.e. in the range(20.5,32.5). As we need least number within this range, we will choose the 7 values in the range 30-39 such that they are at least 33.
So, minimum is 2.
b)Now we employ similar strategy for maximizing
Choose all 4 items in the range 10-19 as 19
Choose all 6 items in the range 20-29 in such a way that median is 24.5. For instance, consider 24,24,24,24,24,25
Choose all 7 items in the range 30-39 as 30
As the median is 24.5, 4 items with values 19, 6 items with values(24,25) and 7 items with values 30 are within the 6 pages from median(24.5). Total = 4+6+7 = 17. This is the maximum we can get.
Btw, what is the source of this?
Last edited by
Frankenstein on Sun Aug 07, 2011 10:00 pm, edited 1 time in total.
Cheers!
Things are not what they appear to be... nor are they otherwise
