Question

the owner of a manufacturing plant employs eighty people. as part of their personnel file, she asked each one to record to the nearest one - tenth of a mile the distance they travel one way from home to work. the distances for a random sample of six employees are listed below: 26, 32, 29, 16, 45, 19. find the variance for the given data. a. 5043.6 mi² b. 18.9 mi² c. 107.0 mi² d. 15.8 mi²

Explanation:

Step1: Identify the data set

The data points are \( 26, 32, 29, 16, 45, 19 \).

Step2: Calculate the mean (\(\bar{x}\))

The formula for the mean is \(\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n}\), where \(n = 6\) (number of data points).
\[
\sum_{i = 1}^{6} x_{i}=26 + 32+29 + 16+45 + 19=167
\]
\[
\bar{x}=\frac{167}{6}\approx27.8333
\]

Step3: Calculate the squared differences from the mean

For each data point \(x_{i}\), calculate \((x_{i}-\bar{x})^{2}\):

• For \(x = 26\): \((26 - 27.8333)^{2}=(- 1.8333)^{2}\approx3.3611\)
• For \(x = 32\): \((32 - 27.8333)^{2}=(4.1667)^{2}\approx17.3611\)
• For \(x = 29\): \((29 - 27.8333)^{2}=(1.1667)^{2}\approx1.3611\)
• For \(x = 16\): \((16 - 27.8333)^{2}=(-11.8333)^{2}\approx140.0370\)
• For \(x = 45\): \((45 - 27.8333)^{2}=(17.1667)^{2}\approx294.6944\)
• For \(x = 19\): \((19 - 27.8333)^{2}=(-8.8333)^{2}\approx78.0278\)

Step4: Sum the squared differences

\[
\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=3.3611 + 17.3611+1.3611 + 140.0370+294.6944+78.0278 = 534.8425
\]

Step5: Calculate the variance (\(s^{2}\))

Since this is a sample (we have a sample of 6 employees from 80), we use the sample variance formula \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\[
s^{2}=\frac{534.8425}{6 - 1}=\frac{534.8425}{5}=106.9685\approx107.0
\]

Answer:

C. \(107.0\space\text{mi}^2\)