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5 positive integers with a mean,median, range and mode of 6? Please helpp

reemurr96 Feb 20, 2019

#1**0 **

The median is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. In simple terms, it may be thought of as the "middle" value of a data set. For example, in the data set {1, 3, 3, 6, 7, 8, 9}, the median is 6, the fourth number in the sample. The median is a commonly used measure of the properties of a data set in statistics and probability theory.

Nickolas Feb 20, 2019

#2

#9**0 **

Median: place the numbers you are given in value order and find the middle number. When the number of numbers is even, the median is the average of the two middle numbers.

In this case we have 6 numbers, so the median (5) will be the average between the 3rd and 4th number when all the positive integers are put in increasing order. (C+D)=10

The next thing we use is the mode (8), which is the most popular value. In a list of 6 numbers we have either two or three 8s.

Now using a bit of common sense, we can say that C and D cannot both be 5, as that would limit us to have only two 8s as E and F, which would mean 8 could not be the mode. Therefore, let’s try C and D as 4 and 6.

Now we have A,B,4,6,8,8

We know that the mean is 5, so we can set an equation.

(A+B+4+6+8+8)/6 = 5 * times both sides by 6

A+B+4+6+8+8 = 30

A+B+26 = 30

A+B = 30–26

A+B = 4

A and B could both be 2, but that would remove the ‘mode status’ from 8. Therefore, A and B have to be 1 and 3 in that order.

The six numbers are: 1,3,4,6,8,8

bigbrotheprodude Feb 20, 2019

#13**+2 **

Let the ordering of the integers from lowest to highest be a, b, c, d and e

And we know that [a + b + c + d + e ] / 5 = 6

a + b + c + d + e = 30

And the median is 6, so....c = 6

And ... e - a = 6 → e = 6 + a

So....

a + b + c + d + e = 30

a + b + 6 + d +( 6 + a) = 30

2a + b + d = 18

b + d = 18 - 2a

[ b + d ] / 2 = 9 - a

"a" must be odd.....so...."e" must be odd

And b, d must be of the same parity....both even or both odd

And since the mode is 6, either b or d (or both ) are 6

Let a = 1 and b = 6... then d would have to be 10

But e = 7....which means that d > e......not possible

And if d = 6, then b = 10....but b > c ....not possible

So a cannot be 1

Next....let a = 3

[ b + d ] / 2 = 9 - 3

b + d = 12

So....if either b, d = 6....then the other must also = 6

So....

3 + 6 + 6 + 6 + 9 = 30 is true

The mean, mode, range and median = 6

So....the numbers are just as EP found !!!!

CPhill Feb 20, 2019