Question

μ = \boxed{}

1. fill in the table below: fill in the differences of each data value from the mean, then the squared differences.

\

$$\begin{tabular}{|c|c|c|} \\hline $x$ & $x - \\mu$ & $(x - \\mu)^2$ \\\\ \\hline 9 & \\boxed{} & \\boxed{} \\\\ \\hline 16 & \\boxed{} & \\boxed{} \\\\ \\hline 6 & \\boxed{} & \\boxed{} \\\\ \\hline 15 & \\boxed{} & \\boxed{} \\\\ \\hline 11 & \\boxed{} & \\boxed{} \\\\ \\hline & $\\sum (x - \\mu)^2 =$ & \\boxed{} \\\\ \\hline \\end{tabular}$$

1. calculate the population standard deviation ($\sigma$).

$\sigma = \sqrt{\frac{\sum (x - \mu)^2}{n}} = \boxed{}$ (please round your answer to two decimal places)
note: $n$ is the population size or the number of data value in the

Explanation:

Step 1: Calculate the population mean (\(\mu\))

The data values are \(9, 16, 6, 15, 11\). The formula for the population mean is \(\mu=\frac{\sum x}{N}\), where \(N\) is the number of data points.
First, find \(\sum x\): \(9 + 16+6 + 15+11=57\).
There are \(N = 5\) data points. So, \(\mu=\frac{57}{5}=11.4\).

Step 2: Fill in the table for \(x - \mu\) and \((x - \mu)^2\)

• For \(x = 9\):
• \(x-\mu=9 - 11.4=- 2.4\)
• \((x - \mu)^2=(-2.4)^2 = 5.76\)
• For \(x = 16\):
• \(x-\mu=16 - 11.4 = 4.6\)
• \((x - \mu)^2=(4.6)^2=21.16\)
• For \(x = 6\):
• \(x-\mu=6 - 11.4=-5.4\)
• \((x - \mu)^2=(-5.4)^2 = 29.16\)
• For \(x = 15\):
• \(x-\mu=15 - 11.4 = 3.6\)
• \((x - \mu)^2=(3.6)^2 = 12.96\)
• For \(x = 11\):
• \(x-\mu=11 - 11.4=-0.4\)
• \((x - \mu)^2=(-0.4)^2=0.16\)

Step 3: Calculate \(\sum(x - \mu)^2\)

Sum up the squared differences: \(5.76+21.16 + 29.16+12.96+0.16=69.2\)

Step 4: Calculate the population standard deviation (\(\sigma\))

The formula for population standard deviation is \(\sigma=\sqrt{\frac{\sum(x - \mu)^2}{N}}\).
We know \(\sum(x - \mu)^2 = 69.2\) and \(N = 5\).
First, calculate \(\frac{\sum(x - \mu)^2}{N}=\frac{69.2}{5}=13.84\)
Then, \(\sigma=\sqrt{13.84}\approx3.72\)

Final Answers:

1. \(\mu=\boldsymbol{11.4}\)
2. Table:

\(x\)	\(x-\mu\)	\((x - \mu)^2\)
16	4.6	21.16
6	-5.4	29.16
15	3.6	12.96
11	-0.4	0.16
	\(\sum(x - \mu)^2=\)	69.2

1. \(\sigma\approx\boldsymbol{3.72}\)

Answer:

Step 1: Calculate the population mean (\(\mu\))

The data values are \(9, 16, 6, 15, 11\). The formula for the population mean is \(\mu=\frac{\sum x}{N}\), where \(N\) is the number of data points.
First, find \(\sum x\): \(9 + 16+6 + 15+11=57\).
There are \(N = 5\) data points. So, \(\mu=\frac{57}{5}=11.4\).

Step 2: Fill in the table for \(x - \mu\) and \((x - \mu)^2\)

• For \(x = 9\):
• \(x-\mu=9 - 11.4=- 2.4\)
• \((x - \mu)^2=(-2.4)^2 = 5.76\)
• For \(x = 16\):
• \(x-\mu=16 - 11.4 = 4.6\)
• \((x - \mu)^2=(4.6)^2=21.16\)
• For \(x = 6\):
• \(x-\mu=6 - 11.4=-5.4\)
• \((x - \mu)^2=(-5.4)^2 = 29.16\)
• For \(x = 15\):
• \(x-\mu=15 - 11.4 = 3.6\)
• \((x - \mu)^2=(3.6)^2 = 12.96\)
• For \(x = 11\):
• \(x-\mu=11 - 11.4=-0.4\)
• \((x - \mu)^2=(-0.4)^2=0.16\)

Step 3: Calculate \(\sum(x - \mu)^2\)

Sum up the squared differences: \(5.76+21.16 + 29.16+12.96+0.16=69.2\)

Step 4: Calculate the population standard deviation (\(\sigma\))

The formula for population standard deviation is \(\sigma=\sqrt{\frac{\sum(x - \mu)^2}{N}}\).
We know \(\sum(x - \mu)^2 = 69.2\) and \(N = 5\).
First, calculate \(\frac{\sum(x - \mu)^2}{N}=\frac{69.2}{5}=13.84\)
Then, \(\sigma=\sqrt{13.84}\approx3.72\)

Final Answers:

1. \(\mu=\boldsymbol{11.4}\)
2. Table:

\(x\)	\(x-\mu\)	\((x - \mu)^2\)
16	4.6	21.16
6	-5.4	29.16
15	3.6	12.96
11	-0.4	0.16
	\(\sum(x - \mu)^2=\)	69.2

1. \(\sigma\approx\boldsymbol{3.72}\)