Certainly! Let's go through the problem step-by-step.
### Part (a)(i)
Show that the mean of [tex]\( x \)[/tex] is 16.5.
Given:
- Sum of squares of deviations from the mean: [tex]\( S_{xx} = 133 \)[/tex]
- Sum of the squares of the values: [tex]\( \sum x^2 = 2311 \)[/tex]
- Number of locations, [tex]\( n = 8 \)[/tex]
Let's find the mean of [tex]\( x \)[/tex].
First, calculate the sum of the values, [tex]\(\sum x\)[/tex], using the mean:
[tex]\[
\text{mean}_x = \frac{\sum x}{n}
\][/tex]
Given that the mean is 16.5, we can write:
[tex]\[
16.5 = \frac{\sum x}{8}
\][/tex]
Solving for [tex]\(\sum x\)[/tex]:
[tex]\[
\sum x = 16.5 \times 8 = 132
\][/tex]
Thus, using the given mean, we determine that [tex]\(\sum x = 132\)[/tex].
Let's recalculate the mean to confirm:
[tex]\[
\text{mean}_x = \frac{\sum x}{n} = \frac{132}{8} = 16.5
\][/tex]
So, we've shown that the mean [tex]\( x \)[/tex] is indeed 16.5.
### Part (a)(ii)
Find the standard deviation of [tex]\( x \)[/tex].
Standard Deviation Formula:
First, we need to find the variance before calculating the standard deviation. The formula for the sample variance is:
[tex]\[
s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1}
\][/tex]
Given:
- [tex]\(\sum x^2 = 2311 \)[/tex]
- [tex]\(\sum x = 132 \)[/tex]
- [tex]\( n = 8 \)[/tex]
Substitute the values into the formula:
[tex]\[
s^2 = \frac{2311 - \frac{132^2}{8}}{8-1}
\][/tex]
Calculate the term [tex]\(\frac{(\sum x)^2}{n}\)[/tex]:
[tex]\[
\frac{132^2}{8} = \frac{17424}{8} = 2178
\][/tex]
Now, calculate the variance [tex]\( s^2 \)[/tex]:
[tex]\[
s^2 = \frac{2311 - 2178}{7} = \frac{133}{7} = 19
\][/tex]
The standard deviation is the square root of the variance:
[tex]\[
s = \sqrt{19} \approx 4.358898943540674
\][/tex]
So, the standard deviation of [tex]\( x \)[/tex] is approximately [tex]\( 4.358898943540674 \)[/tex].
