Chapter 11.3, Problem 6Q

To determine

Make a box-and-whisker plot.

Answer to Problem 6Q

Median = $40$

  $Q_1=25$

  $Q_3=50$

Max = $60$

Min = $20$

Explanation of Solution

Given Information: Minutes of violin practice: $20,50,60,40,40,30,60,40,50,20,20,35$

Formula Used: Median = $\frac{{\left(\frac{n}{2}\right)}^{th}+{\left(\frac{n}{2}+1\right)}^{th}}{2}term$

Calculation:

Minutes of violin practice: $20,50,60,40,40,30,60,40,50,20,20,35$

As, in ascending order

Minutes of violin practice: $20,20,20,30,35,40,40,40,50,50,60,60$

• Median

Median for even numbers = $\frac{{\left(\frac{n}{2}\right)}^{th}+{\left(\frac{n}{2}+1\right)}^{th}}{2}term$

So,

Median = $\frac{{\left(\frac{12}{2}\right)}^{th}+{\left(\frac{12}{2}+1\right)}^{th}}{2}term$

Now,

Median = $\frac{{\left(6\right)}^{th}+{\left(7\right)}^{th}}{2}term$

As, ${\left(6\right)}^{th}term=40$ and ${\left(7\right)}^{th}term=40$

So,

Median = $\frac{40+40}{2}$

Hence,

Median = $40$

• The first quartile, denoted by  $Q_1$, is the median of the lower half of the data set.

So,

  $Q_1=\frac{20+30}{2}$

Hence,

  $Q_1=25$

• The first quartile, denoted by  $Q_3$, is the median of the upper half of the data set.

So,

  $Q_3=\frac{50+50}{2}$

Hence,

  $Q_3=50$

• Max = $60$
• Min = $20$