cryptoaops
Mar 2, 2021

#1**+1 **

Not sure about this... But here's my rational

We know that we have \(15\) scores of \(10\).

Since the mode is \(7\) we need to have \(16\) scores of \(7\) (and no more)

The median is \(9\), so the \(20\) and \(21st\) scores must be \(9...\) to keep the average as small as possible.

And the other \(3\) scores need to be as small as possible so let them \(=1\)

We have

\(1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7 ,7, 7, 9 ,9, 9, 9, 9, 9,10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10\)

\(\text{Median} = 9\)

\(\text{Mode} = 7\)

\(\text{Mean} = 7.975 = 8\)

Credit to CPhill; https://web2.0calc.com/questions/basic-stats_2#r1

P.S. I changed up CPhill's answer a little, so please give me a little credit also!

cryptoaopsDec 28, 2020