Question

a. find the deviation from the mean for each of the data items
b. find the sum of the deviations in part (a).

question 2

question 3

question 4

question 5

question 6

(simplify your answers.)

Explanation:

Step1: Calculate the mean

First, we find the mean ($\bar{x}$) of the data set \(\{18, 30, 42, 72, 108, 162\}\). The formula for the mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n\) is the number of data points and \(x_{i}\) are the data values.

\(n = 6\)

\(\sum_{i=1}^{6}x_{i}=18 + 30+42 + 72+108+162\)

\(18+30 = 48\); \(48+42=90\); \(90 + 72=162\); \(162+108 = 270\); \(270+162=432\)

\(\bar{x}=\frac{432}{6}=72\)

Step2: Calculate deviation for each data point

The deviation from the mean for a data point \(x_{i}\) is \(x_{i}-\bar{x}\)

• For \(x = 18\): \(18-72=- 54\)
• For \(x = 30\): \(30 - 72=-42\)
• For \(x = 42\): \(42-72=-30\)
• For \(x = 72\): \(72 - 72 = 0\)
• For \(x = 108\): \(108-72 = 36\)
• For \(x = 162\): \(162-72=90\)

Step3: Calculate the sum of deviations (for part b)

We sum up the deviations we found in step 2:

\((-54)+(-42)+(-30)+0 + 36+90\)

\((-54-42-30)+(0 + 36+90)\)

\((-126)+(126)=0\)

Answer:

a. The deviations from the mean are: \(-54\), \(-42\), \(-30\), \(0\), \(36\), \(90\) (corresponding to data points \(18\), \(30\), \(42\), \(72\), \(108\), \(162\) respectively)

b. The sum of the deviations is \(0\)