Find the mean, the median and the mode of the data set 3, 7, 5, 9, 5, 2.

Step-by-step solution

1. 1

   Mean: add every value, then divide by how many there are. The mean uses the raw order, so no sorting is needed:

   $3+7+5+9+5+2=31,\qquad n=6,$

   $\bar{x}=\frac{31}{6}=5.1\overline{6}\approx 5.1667.$

   Leaving the answer as the exact fraction $\tfrac{31}{6}$ is better practice than rounding, because $31$ is not divisible by $6$ and any decimal is an approximation.

2. 2

   Sort the data before touching the median. The median is defined on the ordered list, and this is the step people skip:

   $2,\;3,\;5,\;5,\;7,\;9.$

   Reading the median off the unsorted list would give $\tfrac{5+9}{2}=7$, which is wrong.

3. 3

   Median: with an even count, average the two middle values. For $n=6$ the middle positions are the 3rd and 4th:

   $\text{median}=\frac{x_{3}+x_{4}}{2}=\frac{5+5}{2}=5.$

   In general the positions are $\tfrac{n}{2}$ and $\tfrac{n}{2}+1$ when $n$ is even, and the single position $\tfrac{n+1}{2}$ when $n$ is odd. Here both middle values happen to be $5$, so the median is exactly $5$ with no fractional part.

4. 4

   Mode: tally the frequencies. Counting occurrences: $2$ appears once, $3$ once, $5$ twice, $7$ once, $9$ once. The highest frequency is $2$, achieved only by $5$, so the mode is

   $\text{mode}=5.$

   A data set can have no mode (all frequencies equal) or several modes; here there is exactly one.

5. 5

   Sanity-check the three numbers against each other. The mean $5.1667$ sits slightly above the median $5$, which is what you expect from a set with one large outlying value ($9$) pulling the average up. All three statistics land inside the data range $[2,9]$, as they must.