Question

find the variance of the data.
147, 141, 120, 124, 128
\\(\overline{x} = 132\\)
variance (\\(\sigma^2\\)) = ?

Explanation:

Step1: Recall variance formula

The formula for population variance \(\sigma^2\) is \(\sigma^2=\frac{\sum_{i = 1}^{n}(x_i-\overline{x})^2}{n}\), where \(x_i\) are the data points, \(\overline{x}\) is the mean, and \(n\) is the number of data points. Here, \(n = 5\), \(\overline{x}=132\), and the data points are \(147,141,120,124,128\).

Step2: Calculate \((x_i-\overline{x})^2\) for each \(x_i\)

• For \(x_1 = 147\): \((147 - 132)^2=(15)^2 = 225\)
• For \(x_2 = 141\): \((141 - 132)^2=(9)^2=81\)
• For \(x_3 = 120\): \((120 - 132)^2=(- 12)^2 = 144\)
• For \(x_4 = 124\): \((124 - 132)^2=(-8)^2 = 64\)
• For \(x_5 = 128\): \((128 - 132)^2=(-4)^2 = 16\)

Step3: Sum the squared differences

\(\sum_{i = 1}^{5}(x_i-\overline{x})^2=225 + 81+144 + 64+16\)
\(225+81 = 306\); \(306+144 = 450\); \(450+64 = 514\); \(514+16 = 530\)

Step4: Divide by \(n\) to get variance

\(\sigma^2=\frac{530}{5}=106\)

Answer:

\(106\)