# Miguel takes 5 tests. Each score is a whole number between 0 and 100, inclusive. The mean of his scores is 80, the median is 81, and there is just one mode and it is 88. What is the least possible score Miguel could have received on any one test?

Oct 30, 2017

Possible lowest score = 63

Miguel's Scores: 63, 80, 81, 88, 88

#### Explanation:

To simplify thing, the given are the following:
- There are five tests, with scores of 0 to 100
- $M e a n = 80$
- $M e \mathrm{di} a n = 81$
- $M o \mathrm{de} = 88$

Reviewing the meaning of the measures of central tendencies:
- Mean: Average of the scores
- Median: Middle score
- Mode: The score with the highest frequency

So let Miguel's scores arranged in ascending order be: $a , b , c , d , e$

Since we know that the median (the middle score) is 81, and we have a total of five scores, then we know that

$c = 81$

in this case Miguel's scores would be: $a , b , 81 , d , e$

Now, we know that the mode is 88, this means that at least two of Miguel's scores are 88. But as we can see in our given, there may only be two values greater than 81 in order to satisfy the condition $M e \mathrm{di} a n = 81$ hence, we now know that Miguel's scores are:

$a , b , 81 , 88 , 88$

Lastly, in order to get the values of $a$ and $b$, we can use the fact that the $M e a n = 80$. In order to get the least possible score for Miguel, then we can assume that $b$ is the highest number lower than 81 (we will try to make the two scores, $a$ and $b$, as far apart as possible) hence $b = 80$. Note that $b$ could not be 81 since if it is, then the distribution of Miguel's scores would be bimodal (having two modes: $M o \mathrm{de} = 81 , 88$). So now we know that his scores are the following:

$a , 80 , 81 , 88 , 88$

Considering that the $M e a n = 80$, then we know that

$\frac{a + 80 + 81 + 88 + 88}{5} = 80$

So working on this,
$a + 80 + 81 + 88 + 88 = 80 \cdot 5$
$a + 337 = 400$
$a = 400 - 337$
$a = 63$

Hence Miguel's scores are: 63, 80, 81, 88, 88

Checking if all conditions are satisfied:
$63 , 80 , 81 , 88 , 88$

$M e a n = \frac{63 + 80 + 81 + 88 + 88}{5}$
$M e a n = \frac{400}{5}$
$M e a n = 80$

$M e \mathrm{di} a n = 81$
(since it is the middle score)

$M o \mathrm{de} = 88$
(since two of Miguel's scores are 88, and no other score has a frequency of 2 or more)