QUESTION IMAGE
Question
the following are the ages of 18 history teachers in a school district. 24, 25, 26, 29, 29, 31, 33, 37, 42, 44, 45, 46, 47, 49, 49, 52, 56, 57. notice that the ages are ordered from least to greatest. make a box - and - whisker plot for the data.
Step1: Find minimum and maximum
The minimum is \(24\) and maximum is \(57\) from the data set.
Step2: Calculate median
For \(n = 18\) (even), \(Q_2=\frac{42 + 44}{2}=43\).
Step3: Calculate first quartile
For lower half (\(n_1 = 9\) odd), \(Q_1\) is 5th value of lower - half data, \(Q_1 = 29\).
Step4: Calculate third quartile
For upper half (\(n_2 = 9\) odd), \(Q_3\) is 5th value of upper - half data, \(Q_3=49\).
Step5: Draw the plot
Place dots for min and max, draw box from \(Q_1\) to \(Q_3\) and vertical line at \(Q_2\).
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
To create a box - and - whisker plot, we need to find the five - number summary: minimum, first quartile ($Q_1$), median ($Q_2$), third quartile ($Q_3$), and maximum.
- Minimum and Maximum:
- The data set is \(24,25,26,29,29,31,33,37,42,44,45,46,47,49,49,52,56,57\).
- The minimum value is \(24\) and the maximum value is \(57\).
- Median ($Q_2$):
- Since \(n = 18\) (an even number), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+ 1)\)th ordered data values.
- \(\frac{n}{2}=\frac{18}{2}=9\) and \(\frac{n}{2}+1 = 10\).
- The 9th value is \(42\) and the 10th value is \(44\). So, \(Q_2=\frac{42 + 44}{2}=43\).
- First Quartile ($Q_1$):
- Consider the lower half of the data (the first 9 values: \(24,25,26,29,29,31,33,37,42\)).
- Since \(n_1 = 9\) (an odd number), the first - quartile is the \(\frac{n_1 + 1}{2}\)th value.
- \(\frac{9+1}{2}=5\)th value. So, \(Q_1 = 29\).
- Third Quartile ($Q_3$):
- Consider the upper half of the data (the last 9 values: \(44,45,46,47,49,49,52,56,57\)).
- Since \(n_2=9\) (an odd number), the third - quartile is the \(\frac{n_2 + 1}{2}\)th value.
- \(\frac{9 + 1}{2}=5\)th value. So, \(Q_3=49\).
On the number line:
- Place a dot at the minimum value \(24\) (left - most whisker end).
- Place a dot at the maximum value \(57\) (right - most whisker end).
- Draw a box from \(Q_1 = 29\) to \(Q_3=49\).
- Draw a vertical line inside the box at the median \(Q_2 = 43\).