Question 20536: I need to construct a set of twelve numbers with a mean of 7 a median of 6 and a mode of 8. Can you help.
Teri
Found 2 solutions by venugopalramana, AnlytcPhil:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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..
Answer by AnlytcPhil(1807)  (Show Source):
You can put this solution on YOUR website! I suggest starting with all 7's, so the mean will be 7
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
We can add anything to any of these numbers and as long as we subtract the same
away as the total amount we add on, the mean will remain 7, for the sum will
remain the same. But let's agree to do it so that the 12 numbers will remain
in ascending order from small to large.
To make the median 6, we must make the middle two numbers have a mean of 6.
The easiest way is to make them both 6. But to do this and keep ascending
order we'll have to subtract 1 from each of the first seven 7's. So we'll need
to add 7 among the last 5 7's, trying to make as many 8's as possible since the
mode must be 8. The most number of 8's we can make is four, by adding 1 to
each of the 7's on the right except the last one, to which we must add 3.
6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 10
But there are only four 8's, and seven 6's, so we must reduce the number of 6's
below four. So we subtract 1 from each of the first four 6's, and add 4 to the
10 on the far right, giving
5, 5, 5, 5, 6, 6, 6, 8, 8, 8, 8, 14
But this makes four 5's, causing a tie between the 5's and the 8's for being
the mode, so we subtract 1 from the first 5 and add one to the 14 on the far
right, giving
4, 5, 5, 5, 6, 6, 6, 8, 8, 8, 8, 15
That's one answer. There are lots of other possible answers, such as
2, 4, 4, 5, 5, 5, 7, 8, 8, 8, 8, 20
Edwin
AnlytcPhil@aol.com
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