Question 20536
there are 12 numbers..say x1,x2,x3,.....x11,x12 when arranged in increasing order.
mean is 7.so sum of numbers by /total no.of numbers=7=(x1+x2+....+x11+x12)/12
hence sum of numbers =7*12=84=(x1+x2+....+x11+x12)
median is 6..that is the middle term is 6..but total number of terms is 12 so there would be 2 middle terms...namely x6 and x7...by convention in such cases their average is taken as median..so (x6+x7)/2=6...or....x6+x7=2*6=12.
mode is 8 ...that is the frequency of occurence of mode or number 8 should be maximum...
now there are no unique answers for this... we are just to give any set of 12 numbers satisfying the above 3 conditions.
say one set of numbers could be then ....
   let us put median as 6 ..by taking both x6 and x7 as 6 each...
now 8 should occur maximm numer of times ...so let x8=x9=x10=x11=8..here 8 occurs 4 times.
let us take x12=16 say...
total of numbers we have taken so far is 6+6+8*4+16=60
now take x1 to x5 so that thy add up to  84-60=24...since total should be 84 to get mean 7 as given above...we only have to see that no number should occur more than 4 times as mod of 8 ,we took with frequency of 4 which should be the highest...
so let us take x1=4,x2=4,x3=5,x4=5,x5=6..which satisfies all the above conditions
so the set of numbers in increasing order are 4,4,5,5,6,6,6,8,8,8,8,16....
their mean is (4+4+5+5+6+6+6+8+8+8+8+16)/12=84/12=7
median is 6 ,since middle most number /numers are 6.
mode is 8 as 8 occurs 4 times ,the maximum frequency..