Answer with explanation:
A certain list consists of 3 different numbers.
Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?
Let the three numbers be x, y, and z, where x < y < z.
Now, the median will be y and the average will be
![(x+y+z)/3](https://img.qammunity.org/2020/formulas/mathematics/high-school/wbntkolrw0coqjn9r2zmtvace45yfecpwu.png)
We have to tell if
or
![2y=x+z](https://img.qammunity.org/2020/formulas/mathematics/high-school/rm7crwj9qi110pq21xomrv77gvwo322otg.png)
(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median.
Range is the largest number minus the smallest number of the set,
Here range = z - x.
We are given in the statement that
![z-x=2(z-y)](https://img.qammunity.org/2020/formulas/mathematics/high-school/xunsftluwra11x4kwajbxovd6e1cs19pvm.png)
Solving this we get;
![z-x=2z-2y](https://img.qammunity.org/2020/formulas/mathematics/high-school/fddhiwnxnube6gcc0y8ehhv3ag4fj8p6b9.png)
or
So, this condition is fulfilled.
(2) The sum of the 3 numbers is equal to 3 times one of the numbers.
The sum of 3 numbers cannot be equal to 3 times the smallest number or 3 times largest number as the given numbers are distinct and they cannot be equal to mean. So, median = mean.
![x+y+z=3y](https://img.qammunity.org/2020/formulas/mathematics/high-school/2iige33i3qu9w3o691gyxu09blcupjeihu.png)
or
![x+z=2y](https://img.qammunity.org/2020/formulas/mathematics/high-school/40szom90c4l9b6wpgipukp9npeyg2bkdoe.png)
So, this condition is fulfilled.