Basic statistics

A statistic is a a quantity calculated from a set of data. Useful statistics help describe the characteristics of a data set. This lesson will introduce three basic statistics: the mean, median, and mode.

Mean

A mean is the average of a set of numbers. The mean is calculated by summing all of the numbers then dividing this total by the amount of numbers in the data set. To demonstrate, we’ll calculate the mean of the following data set. This data set contains the test scores of a group of $ 15 $ test takers.

$\begin{align*} \text{Scores: } & 55,68,71,72,72,72,76,80, \\ & 84,84,88,90,90,95,98 \end{align*}$

Adding up all of these scores gives us $ 1,195 $, therefore we can calculate the mean test score as $ 1,195/15=79.7 $.

Try calculating the mean of this data set: $ 16,19,19,20,21,25 $. You should get $ 20 $.

Median

If you sort a data set in ascending or descending order, the number in the middle is the median. The data set of test scores from above is already sorted. The middle number in the list is $ 80 $, which is the median.

The second data set is also sorted, but it has an even number of data points so their is no middle number in the list. When this occurs, the median is calculated by averaging the two ‘middle numbers,’ in this case: $ 19 $ and $ 20 $. Thus our median is: $ (19+20)/2=19.5 $.

Data sets will not always be written in order as the two above were. Try computing the median of the following data set:

$\text{Points: } 8,5,12,23,8,12$

You should get a median of $ (8+12)/2=10 $.

Mode

The mode is the most frequently occurring point in a data set. In the $ Scores $ data set our mode is $ 72 $, because 72 occurs 3 times in the data set–more than any other number. It is possible for a data set to have two modes (bi-modal). The $ Points $ data set above has two modes: 8 and 12. When a data set has more than two modes, we usually say there is no mode. The following data set gives an example.

$\text{Ages: } 2,7,16,11,8,19$

Each data point only occurs once, so there is ‘no mode.’

Practice questions

For the following data sets calculate the mean, median, and modes

$ \text{Weights: } 118,190,145,230,130,145,155 $
$ \text{Rankings: } 9,88,23,43,52,31 $
$ \text{Speeds: } 5.00,4.82,4.47,4.78,4.40,5.20,4.82,4.51 $

Solve

Mrs. Ryan knows the average score of her class on a recent test was 80, but she has misplaced one student’s test. The scores for the other students are: 69,74,76, and 85. What score must be on the missing test?
The average of a data set is 18.3. There are 30 points in the set. If the points 19.8 and 20 are added to the data set, what will the new mean be?

Solutions

$ \begin{align*} & \text{Mean: }1,113/7=159 \\ & \text{Median: }145 \\ & \text{Mode: }145 \end{align*} $
$ \begin{align*} & \text{Mean: }246/6=41 \\ & \text{Median: }(31+43)/2=37 \\ & \text{Mode: None} \end{align*} $
$ \begin{align*} & \text{Mean: }38.00/8=4.75 \\ & \text{Median: }(4.78+4.82)/2=4.80 \\ & \text{Mode: }4.82 \end{align*} $
Let $ x $ represent the missing test score. Using the definition of a mean, the following equation must hold: $$ \frac{x+69+74+76+85}{5}=80 $$ Solving for $ x $, we get: $$ \begin{align*} & \frac{x+304}{5}=80 \\[5pt] & \rightarrow x+304=400 \\[5pt] & \rightarrow x=96 \end{align*} $$
Adding two points creates a new data set of $ 32 $ points. Since the mean of the original $ 30 $ points is $ 18.3 $, these $ 30 $ points must sum to $ 30\cdot18.3 $, because $ (30\cdot18.3)/30=18.3 $. So the mean of our new data set will be: $$ \frac{30\cdot18.3 +19.8+20}{32}=\frac{588.8}{32}=18.4 $$

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