Answer:
[tex]\sigma_1 = 0.08[/tex] --- Location 1
[tex]\sigma_2 = 0.34[/tex] --- Location 2
Step-by-step explanation:
Given
See attachment for the given data
Required
The standard deviation of each location
For location 1
First, calculate the mean
[tex]\bar x_1 =\frac{\sum x}{n}[/tex]
So, we have:
[tex]\bar x_1 =\frac{30.40+30.20+30.30+30.40+30.30}{5}[/tex]
[tex]\bar x_1 =\frac{151.60}{5}[/tex]
[tex]\bar x_1 =30.32[/tex]
The standard deviation is calculated as:
[tex]\sigma_1 = \sqrt{\frac{\sum(x - \bar x_1)^2}{n-1}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{(30.40 - 30.32)^2+(30.20 - 30.32)^2+(30.30 - 30.32)^2+(30.40 - 30.32)^2+(30.30 - 30.32)^2}{5-1}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{0.028}{4}}[/tex]
[tex]\sigma_1 = \sqrt{0.007}[/tex]
[tex]\sigma_1 = 0.08[/tex]
For location 2
First, calculate the mean
[tex]\bar x_2 =\frac{\sum x}{n}[/tex]
So, we have:
[tex]\bar x_2 =\frac{30.10+30.90+30.20+30.70+30.30}{5}[/tex]
[tex]\bar x_2 =\frac{152.2}{5}[/tex]
[tex]\bar x_2 =30.44[/tex]
The standard deviation is calculated as:
[tex]\sigma_2 = \sqrt{\frac{\sum(x - \bar x_2)^2}{n-1}}[/tex]
[tex]\sigma_2 = \sqrt{\frac{(30.10-30.44)^2+(30.90-30.44)^2+(30.20-30.44)^2+(30.70-30.44)^2+(30.30-30.44)^2}{5-1}}[/tex]
[tex]\sigma_2 = \sqrt{\frac{0.472}{4}}[/tex]
[tex]\sigma_2 = \sqrt{0.118}[/tex]
[tex]\sigma_2 = 0.34[/tex]
