Question

1. consider the following set of data:

2.0, 12.8, 3.2, 4.0, 2.8, 6.6, 5.8, 6.6, 2.8, 2.8
11a complete the table.
lower extreme 2.0
lower quartile (q₁) 2.8
median 3.6
upper quartile (q₃) 6.6
upper extreme 12.8
11b determine the outlier.
outlier = enter your next step here

Explanation:

Step1: Recall the outlier formula

To find outliers, we use the interquartile range (IQR) method. The formula for identifying outliers is:

• Lower bound: \( Q_1 - 1.5 \times \text{IQR} \)
• Upper bound: \( Q_3 + 1.5 \times \text{IQR} \)
• Any data point below the lower bound or above the upper bound is an outlier.

First, calculate the IQR: \( \text{IQR} = Q_3 - Q_1 \)

From the table, \( Q_1 = 2.8 \) and \( Q_3 = 6.6 \)
So, \( \text{IQR} = 6.6 - 2.8 = 3.8 \)

Step2: Calculate the lower and upper bounds

• Lower bound: \( Q_1 - 1.5 \times \text{IQR} = 2.8 - 1.5 \times 3.8 = 2.8 - 5.7 = -2.9 \)
• Upper bound: \( Q_3 + 1.5 \times \text{IQR} = 6.6 + 1.5 \times 3.8 = 6.6 + 5.7 = 12.3 \)

Step3: Check each data point

The data set is: \( 2.0, 12.8, 3.2, 4.0, 2.8, 6.6, 5.8, 6.6, 2.8, 2.8 \)

Check each value:

• \( 2.0 \): Between -2.9 and 12.3, not an outlier.
• \( 12.8 \): Greater than 12.3, so it is an outlier.
• \( 3.2 \): Between -2.9 and 12.3, not an outlier.
• \( 4.0 \): Between -2.9 and 12.3, not an outlier.
• \( 2.8 \): Between -2.9 and 12.3, not an outlier.
• \( 6.6 \): Between -2.9 and 12.3, not an outlier.
• \( 5.8 \): Between -2.9 and 12.3, not an outlier.
• \( 6.6 \): Between -2.9 and 12.3, not an outlier.
• \( 2.8 \): Between -2.9 and 12.3, not an outlier.
• \( 2.8 \): Between -2.9 and 12.3, not an outlier.

Answer:

\( 12.8 \)