Question

box-and-whisker plot
level 1:51
make box-and-whisker plots for the given data.

1. 17, 29, 32, 9, 30, 14, 8, 39, 11, 32, 23
2. 58, 67, 44, 72, 51, 42, 60, 46, 69

(includes a number line from 0 to 40 with 10, 20, 30, 40 marked, and spaces for minimum, maximum, q₁, q₂, q₃ for each data set)

Explanation:

Problem 1: Data set - 17, 29, 32, 9, 30, 14, 8, 39, 11, 32, 23

Step 1: Order the data

First, we order the data from smallest to largest: \( 8, 9, 11, 14, 17, 23, 29, 30, 32, 32, 39 \)

Step 2: Find Minimum and Maximum

Minimum is the smallest value: \( 8 \)
Maximum is the largest value: \( 39 \)

Step 3: Find the median (\( Q_2 \))

The number of data points \( n = 11 \) (odd). The median is the middle value (\( \frac{n + 1}{2} = 6 \)-th term).
The 6 - th term in the ordered data is \( 23 \), so \( Q_2 = 23 \)

Step 4: Find \( Q_1 \) (median of the lower half)

The lower half of the data (values below \( Q_2 \)): \( 8, 9, 11, 14, 17 \)
Number of data points in lower half \( n_1 = 5 \) (odd). The median of the lower half is the middle value (\( \frac{5+ 1}{2}=3 \)-rd term).
The 3 - rd term in \( 8, 9, 11, 14, 17 \) is \( 11 \), so \( Q_1 = 11 \)

Step 5: Find \( Q_3 \) (median of the upper half)

The upper half of the data (values above \( Q_2 \)): \( 29, 30, 32, 32, 39 \)
Number of data points in upper half \( n_2 = 5 \) (odd). The median of the upper half is the middle value (\( \frac{5 + 1}{2}=3 \)-rd term).
The 3 - rd term in \( 29, 30, 32, 32, 39 \) is \( 32 \), so \( Q_3 = 32 \)

Problem 2: Data set - 58, 67, 44, 72, 51, 42, 60, 46, 69

Step 1: Order the data

First, we order the data from smallest to largest: \( 42, 44, 46, 51, 58, 60, 67, 69, 72 \)

Step 2: Find Minimum and Maximum

Minimum is the smallest value: \( 42 \)
Maximum is the largest value: \( 72 \)

Step 3: Find the median (\( Q_2 \))

The number of data points \( n=9 \) (odd). The median is the middle value (\( \frac{n + 1}{2}=5 \)-th term).
The 5 - th term in the ordered data is \( 58 \), so \( Q_2 = 58 \)

Step 4: Find \( Q_1 \) (median of the lower half)

The lower half of the data (values below \( Q_2 \)): \( 42, 44, 46, 51 \)
Number of data points in lower half \( n_1 = 4 \) (even). The median of the lower half is the average of the two middle values (\( \frac{4}{2}=2 \)-nd and \( \frac{4}{2}+ 1 = 3 \)-rd terms).
The 2 - nd term is \( 44 \) and the 3 - rd term is \( 46 \). So \( Q_1=\frac{44 + 46}{2}=\frac{90}{2}=45 \)

Step 5: Find \( Q_3 \) (median of the upper half)

The upper half of the data (values above \( Q_2 \)): \( 60, 67, 69, 72 \)
Number of data points in upper half \( n_2 = 4 \) (even). The median of the upper half is the average of the two middle values (\( \frac{4}{2}=2 \)-nd and \( \frac{4}{2}+ 1=3 \)-rd terms).
The 2 - nd term is \( 67 \) and the 3 - rd term is \( 69 \). So \( Q_3=\frac{67+69}{2}=\frac{136}{2} = 68 \)

Answer:

(Problem 1):
Minimum: \( 8 \)
Maximum: \( 39 \)
\( Q_1 \): \( 11 \)
\( Q_2 \): \( 23 \)
\( Q_3 \): \( 32 \)