Draw a Box and Whisker Plot Worksheet
A box and whisker plot (or box plot ) is a graph that displays the information distribution by using five numbers. Those five numbers are: the minimum , first (lower) quartile , median , third (upper) quartile and maximum . Think that we divers median in the lesson Way and median and quartiles in the lesson Quantiles.
Interpreting box and whisker plots
Example ane:Find the range, the interquartile range and the median of the information in the box plot beneath.
Solution:
Retrieve that we divers range and interquartile range in the lesson Other measures of dispersion.
Since the minimum value of the given data is $5$ and maximum $50$, the range is $R_{10} = 50 – five = 45$.
The lower quartile is $xv$ and the upper quartile is $35$. Therefore, the interquartile range is $I_{Q} = 35 – 15 = 20$. Really, the interquartile range represents the length of the box.
The median is plain $25$.
Instance ii:The following data represents the number of sold items of some shop in one 60 minutes during one calendar week:
$$7, thirteen, 5, 17, 26, xx, x.$$
Which of the box plots below represents the given information?
a)
b)
c)
Solution:
First we need to order the information points from smallest to largest:
$$five, 7, x, thirteen, 17, xx, 26.$$
The minimum value is $5$ and the maximum $26$. Therefore, a) is certainly not the answer since for box plot in a) the maximum value is $twenty$. Permit's summate the median of the given data.
Since the number of information points is odd, we have:
$$M_{east} = 13.$$
As we tin can see, the median in the box plot in b) is $M_{e} = 13$. Furthermore, median in the box plot in c) is $M_{e} = 15$.
In conclusion, the right answer is b).
Creating box and whisker plots
Example 3: Martha threw the die $20$ times and got these results:
$$half-dozen \ 3 \ 3 \ 6 \ 3 \ 5 \ 6 \ 1 \ 4 \ 6 $$
$$three \ 5 \ 5 \ 2 \ 2 \ 2 \ two \ three \ 2 \ iii.$$
Depict a box plot.
Solution:
The first thing we need to do is to social club the data from smallest to largest:
$$1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 5, five, 5, 6, vi, 6, half dozen.$$
Furthermore, we need to calculate the median. Since the number of information points is fifty-fifty, we have
$$Me = \frac{x_{10}+x_{11}}{two} = \frac{three + iii}{ii} = iii.$$
Later that nosotros have to calculate the quartiles.
As we mentioned in the lesson Quantiles, the lower quartile is the middle value of the information points on the left side of the median. Furthermore, the upper quartile is the middle value of the data points to the right of the median. In other words, the lower quartile is
$$Q_{one} = \frac{2 + two}{2} = 2,$$
while the upper quartile is
$$Q_{three} = \frac{5 + five}{2} = 5.$$
The minimum is the smallest data point, i.e. $1$. The maximum is the largest data point, i.eastward. $half-dozen$.
The side by side step is to scale an appropriate axis for obtained $v$ numbers.
Then we demand to describe a box from $Q_{one}$ to $Q_{3}$ and put a vertical line through the median. Furthermore, we take to draw "whiskers". Those are the lines which extend parallel with the scale from the box. In other words, whisker goes from $Q_{1}$ to the minimum and from $Q_{3}$ to the maximum.
Finally, our box plot is:
Eastwardxample 4: The post-obit information are the weights (in kg) of $twenty$ students:
$$64, 50, 53, 89, 54, 55, 57, 75, 57, 92, 58, 61, 63, 66, 67, 70, 76, 85, 88, 95.$$
Draw a box plot.
What percentage of students accept more than $80$ kg? What percentage have less than $65$ kg?
Solution:
Past ordering the given prepare, we obtain
$$50, 53, 54, 55, 57, 57, 58, 61, 63, 64, 66, 67, 70, 75, 76, 85, 88, 89, 92, 95.$$
The minimum value is $50$, while the maximum value is $95$.
Now we can summate the median:
$$M_{e} = \frac{64 + 66}{2} = 65.$$
Furthermore,
$$Q_{one} = \frac{57 + 57}{2} = 57, Q_{3} = \frac{76 + 85}{ii} = 80.v.$$
Box plot is:
$eighty$ kg is an upper quartile $Q_{3}$, so one section of the box plot is greater than $Q_{3}$.
Since each section has $25 \%$ of the given data, we conclude that $25\%$ of the students have more than $80$ kg.
Similarly, $65$ kg is the median of the given data. Therefore, $l\%$ of the students have less than $65$ kg.
Comparison box and whisker plots
Example 5: The box plots beneath show an amount of fourth dimension that men and women spend per day reading.
Using the box plots, reply the questions.
Fourth dimension – men:
Time – women:
a) Approximate the interquartile range for the given box plots.
b) What percentage of men spend more than than $two.5$ hours per twenty-four hours reading? Similarly, what percentage of women spend more than $2.5$ hours per day reading?
Solution:
a) For men:
The lower quartile is approximately $Q_{ane} = i.25$. In other words, $ane.25$ hours per day.
The upper quartile is $Q_{3} = 2.5$ hours per 24-hour interval.
Therefore, the interquartile range is $I_{Q} = 2.5 – ane.25 = one.25$ hours per twenty-four hour period.
For women:
The lower quartile is approximately $Q_{1} = 1.25$ hours per day.
The upper quartile is $Q_{3} = iii$ hours per day.
In other words, the interquartile range is $I_{Q} = 3 – 1.25 = i.75$ hours per twenty-four hours.
b) For men:
$2.5$ hours is an upper quartile $Q_{three}$, so one department of the box plot is greater than $Q_{3}$. Since each section has $25 \%$ of the given data, we conclude that $25\%$ of men spend more than $2.five$ hours per day reading.
For women:
$ii.five$ hours is the median $M_{e}$ of the given data. Similarly as for the first box plot, nosotros conclude that $50\%$ of women spend more than $two.5$ hours per 24-hour interval reading.
murray-priorcabou1994.blogspot.com
Source: https://mathemania.com/lesson/box-and-whisker-plot/