Q5. - Algebra 1

Question

Use each set of data to make a stem-and-leaf plot and a box-and-whisker plot. Describe how the outliers affect the quartile points.

{31,30,28,26,22,34,26,31,47,32,18,33,26,23,18}

Step-by-Step Solution

Answer

The required stem-and-leaf table for the given data is:

The required box and whiskers plot for the given data is:

The outlier lies far away from the quartile points.

Step-1. Apply the concept of stem and leaf plot.

A Stem and Leaf Plot is a special table where each data value is split into a "stem" (the first digit or digits) and a "leaf" (usually the last digit).

Step-2. Apply the concept of the interquartile range (IQR), median(Q2), lower quartile(Q1), and upper quartile(Q3).

• The interquartile range is $IQR=Q_3-Q_1$
• The median is the middle value when a data set is ordered from least to greatest. It is the second quartile (Q2) and divides the data into two half.
• The lower quartile Q1 is the median of the lower half of the data. 
• The upper quartile Q3 is the median of the upper half of the data.
• If the size of the data set is odd, include the median when finding the first and third quartiles.
• If the size of the data set is even, the median splits the data set into lower and upper halves.

Step-3. Analyze the data.

Given data:

{31,30,28,26,22,34,26,31,47,32,18,33,26,23,18}

Arrange the data in ascending order:

18,18,22,23,26,26,26,28,30,31,31,32,33,34,47

Step-4. Draw the stem and leaf plot.

The greatest place value is tenth.

So, number 18 will have stem 1 and leaf 8.

Similarly, applies for all the numbers.

Step-5. Calculation of median (Q2).

Given data:

{31,30,28,26,22,34,26,31,47,32,18,33,26,23,18}

Arrange the data in ascending order:

18,18,22,23,26,26,26,28,30,31,31,32,33,34,47

So, the median is average of ${\left(\frac{15+1}{2}\right)}^{th}=8^{th}$ and $7^{th}$ observation

Therefore, the median is $Q_2=28$

Step-6. Calculation of lower quartile(Q1).

Arrange the data in ascending order:

18,18,22,23,26,26,26,28,30,31,31,32,33,34,47

The lower half of the data set is:

18,18,22,23,26,26,26

The lower quartile is the median of lower half of the data set:

$\Rightarrow Q_1=23$

Step-7. Calculation of upper quartile(Q3).

Arrange the data in ascending order:

18,18,22,23,26,26,26,28,30,31,31,32,33,34,47

The upper half of the data set is:

30,31,31,32,33,34,47

The upper quartile is the median of the upper half of the data set:

$\Rightarrow Q_3=32$

Step-8. Calculation of interquartile range.

$\begin{array}{l}IQR=Q_3-Q_1\\ \Rightarrow IQR=32-\mathrm{23.........}[\text{from step 6 and 7]}\\ \Rightarrow IQR=9\end{array}$

Step-9. Check the outliers.

$\begin{array}{l}{Q}_1-1.5\times IQR\text{ and }{Q}_3+1.5\times IQR\\ 23-1.5\times 9\text{ and 32}+1.5\times 9\\ 23-13.5\text{ and 32}+13.5\\ 9.5\text{ and 45}.5\end{array}$

Numbers less than 9.5 or more than 45.5 are outliers.

So, 47 is the outlier.

Step-10. Draw box-and-whisker plot.

Draw a number line that includes least and greatest numbers in the data. Mark the three quartile points, the least number that is not an outlier, and the greatest number that is not an outlier by vertical line segments.

Draw the box and the whiskers. The box goes through the quartiles and outliers are not connected to the box.

From the diagram, we can clearly figure out that outlier 47 lies far away from the quartile points.