Box and Whisker Plots

1. Place the data in ascending order

|12,| |52,| |60,| |60,| |63,| |65,| |65,| |68,| |68,| |70,| |72,| |74,| |75,| |76,| |76,| |80,| |84,| |88,| |92,| |94,| |96,| |98|

2. Separate the data distribution into |\boldsymbol{4}| equal quarters

This distribution has an even number of data values |(22).| Therefore, |Q_2| is between the |2| data at the centre of the distribution and separates it into |2| subgroups of |11| data. |Q_1| and |Q_3| are therefore the data values in the middle of their respective subgroups, so as to create |4| quarters that contain |5| data values each.
||\begin{alignat}{20}&&&\!\!\!\!\boldsymbol{\color{#ec0000}{\underbrace{Q_2}}}\\12,52,60,60,63\ &\color{#3b87cd}{\boxed{\boldsymbol{65}}}\ 65,68,68,70,72\ &&\color{#ec0000}{\Big\vert}\ 74,75,76,76,80\ &&\color{#7cca51}{\boxed{\boldsymbol{84}}}\ 88,92,94,96,98\\&\!\!\;\boldsymbol{\color{#3b87cd}{\overbrace{Q_1}}}&&&&\!\!\;\boldsymbol{\color{#7cca51}{\overbrace{Q_3}}}\end{alignat}||

3. Determine the value of the quartiles

We start by determining the value of the median |(Q_2),| which corresponds to the mean of the 11th and 12th data values.
||\begin{alignat}{20}&&&\!\!\!\!\boldsymbol{\color{#ec0000}{\underbrace{Q_2}}}\\12,52,60,60,63\ &\color{#3b87cd}{\boxed{\boldsymbol{65}}}\ 65,68,68,70,\boldsymbol{72}\ &&\color{#ec0000}{\Big\vert}\ \boldsymbol{74},75,76,76,80\ &&\color{#7cca51}{\boxed{\boldsymbol{84}}}\ 88,92,94,96,98\\\phantom{\boldsymbol{\overbrace{Q_1}}}\end{alignat}||||\boldsymbol{\color{#ec0000}{Q_2}}=\dfrac{72+74}{2}=\boldsymbol{\color{#ec0000}{73}}||
Next, we determine the 1st quartile |(Q_1),| which corresponds to the 6th data value.
||\boldsymbol{\color{#3b87cd}{Q_1}}=\boldsymbol{\color{#3b87cd}{65}}||
Finally, we determine the 3rd quartile |(Q_3),| which corresponds to the 17th data value.
||\boldsymbol{\color{#7cca51}{Q_3}}=\boldsymbol{\color{#7cca51}{84}}||

4. Determine if there are any outliers

First, the interquartile range is calculated.
||\begin{align}IR&=Q_3-Q_1\\&=84–65\\&=19\end{align}||
Next, we verify if the data at the ends of the distribution are outliers.

Therefore, |12| is an outlier.

5. Determine the minimum and maximum

Since |12| is an outlier, it is not considered the minimum value of the distribution.
||\begin{align}x_\text{min}&=52\\x_\text{max}&=98\end{align}||

6. Draw the box and whisker plot

Using a number line and the values calculated in the previous steps, the box and whisker plot is drawn. The outlier |(12)| is represented by a star.

a) Who had the fewest number of sales in one day?

Simply compare the minimum data values |(x_\text{min})| for Karin and Eric.

The person with the fewest number of sales in one day was Karine.

b) Who is the most consistent in terms of sales?

We can compare the range and the interquartile range |({IR})| for Karine and Eric, which will give an idea of the concentration of the data in each distribution.

We see that Karine's range and interquartile range of sales are smaller to Eric's. In other words, compared to Eric, Karine's daily sales are more concentrated around the median, implying that Karine is a more consistent salesperson than Eric.

c) Considering their respective |\boldsymbol{15}| best days, who made the most sales?

To answer this question, we must analyze the data. However, since we do not know all the data values, we assume that they are evenly distributed in each quarter.

The sales data was collected over a period of |30| days. Therefore, we are interested in |50\%| of the days |\left(\dfrac{15}{30}\right)| that Karin and Eric made the most sales. In other words, we need to analyze the 3rd and 4th quarters of each box and whisker plot.

We notice that the median and the maximum of Eric's sales are both higher than those of Karine. Therefore, we can conclude that Eric made the most sales during his best |15| days.

d) Did Karine necessarily make a day with |\boldsymbol{36}| sales?

We note that |36| is the median |(Q_2)| of Karine's distribution. However, the data was collected over a period of |30| days. The median is therefore the mean of the 15th and 16th data values.
||\begin{alignat}{20}&\!\!\!\!\!\!\!\!\!\!\!\boldsymbol{\color{#ec0000}{\underbrace{Q_2=36}}}\\x_\text{min}\ldots15\text{th}\text{ data value}\ &\color{#ec0000}{\Big\vert}\ 16\text{th}\text{ data value}\ldots x_\text{max}\end{alignat}||
For instance, the 15th data value might be |35| and the 16th data value |37.| The median would therefore be calculated as follows:
||\boldsymbol{\color{#ec0000}{Q_2}}=\dfrac{35+37}{2}=36||
We conclude that Karine did not necessarily make a day with 36 sales.

e) For how many days did Eric make between |\boldsymbol{22}| and |\boldsymbol{40}| sales per day?

We are concerned with the number of days in the 2nd quarter, that is, between |22| |(Q_1)| and |40| |(Q_2).| We found earlier that the median |(Q_2)| separates the distribution into 2 subgroups of |15| data values. This means that |Q_1| corresponds to the 8th data value in the distribution.
||\begin{alignat}{20}&&&\!\!\!\!\!\!\!\!\!\!\!\boldsymbol{\color{#ec0000}{\underbrace{Q_2=40}}}\\x_\text{min}\ldots7\text{th}\text{ data value}\ &\color{#3b87cd}{\boxed{\boldsymbol{8}\textbf{th}\textbf{ data value}}}\ 9\text{th}\text{ data value}\ldots15\text{th}\text{ data value}\ &&\color{#ec0000}{\Big\vert}\ 16\text{th}\text{ data value}\ldots x_\text{max}\\&\ \:\boldsymbol{\color{#3b87cd}{\overbrace{Q_1=22}}}\end{alignat}||
Each quarter contains |7| data values. So, unless the data values |22| and |40| repeat, we can conclude that Eric made between |22| and |40| sales during |7| days.