Question

name
alg 1.1 lesson 5 redo calculating measures of center and variability
dataset: 12, 5, 9, 7, 2
mean: 7

1. calculate the mean absolute deviation (mad) for the data
2. calculate the interquartile range (iqr) for the data

Explanation:

1) Calculating Mean Absolute Deviation (MAD)

Step 1: Recall MAD formula

The formula for the mean absolute deviation is \( \text{MAD} = \frac{1}{n}\sum_{i = 1}^{n}|x_{i}-\bar{x}| \), where \( n \) is the number of data points, \( x_{i} \) are the data points, and \( \bar{x} \) is the mean. Here, \( n = 5 \), \( \bar{x}=7 \), and the data points are \( 12, 5, 9, 7, 2 \).

Step 2: Calculate absolute deviations

• For \( x_1 = 12 \): \( |12 - 7|=5 \)
• For \( x_2 = 5 \): \( |5 - 7| = 2 \)
• For \( x_3 = 9 \): \( |9 - 7| = 2 \)
• For \( x_4 = 7 \): \( |7 - 7| = 0 \)
• For \( x_5 = 2 \): \( |2 - 7| = 5 \)

Step 3: Sum the absolute deviations

Sum \( = 5 + 2 + 2 + 0 + 5 = 14 \)

Step 4: Divide by \( n \)

\( \text{MAD}=\frac{14}{5}=2.8 \)

Step 1: Order the data

First, we order the data set: \( 2, 5, 7, 9, 12 \)

Step 2: Find the median (Q2)

Since \( n = 5 \) (odd), the median is the middle value. The middle value is the 3rd term, so \( \text{Median}(Q_2)=7 \)

Step 3: Find Q1 (median of lower half)

The lower half of the data (values below the median) is \( 2, 5 \). The median of this set (since there are 2 values) is the average of the two: \( Q_1=\frac{2 + 5}{2}=3.5 \)

Step 4: Find Q3 (median of upper half)

The upper half of the data (values above the median) is \( 9, 12 \). The median of this set is the average of the two: \( Q_3=\frac{9+12}{2}=10.5 \)

Step 5: Calculate IQR

The interquartile range is \( \text{IQR}=Q_3 - Q_1 \). So, \( \text{IQR}=10.5 - 3.5 = 7 \)

Answer:

The mean absolute deviation (MAD) is \( 2.8 \)

2) Calculating Interquartile Range (IQR)