Let's solve the problem step-by-step:
### (a) Five-number summary
The five-number summary includes:
1. Minimum value
2. First quartile (Q1)
3. Median
4. Third quartile (Q3)
5. Maximum value
Given the data set: [tex]\(\{28, 34, 27, 42, 52, 15\}\)[/tex]
- Minimum: The smallest value in the set is [tex]\(15\)[/tex].
- First quartile (Q1): The 25th percentile of the data is [tex]\(27.25\)[/tex].
- Median: The middle value of the ordered data set is [tex]\(31.0\)[/tex].
- Third quartile (Q3): The 75th percentile of the data is [tex]\(40.0\)[/tex].
- Maximum: The largest value in the set is [tex]\(52\)[/tex].
So, the five-number summary is:
[tex]\[
(15, 27.25, 31.0, 40.0, 52)
\][/tex]
### (b) Mean of the data set
The mean, denoted as [tex]\(\bar{x}\)[/tex], is the average of the data values.
Given the data set, the mean is calculated as:
[tex]\[
\bar{x} = \frac{28 + 34 + 27 + 42 + 52 + 15}{6} = 33.0
\][/tex]
### (c) Sum of the squares of the differences between each data value and the mean
We will use the table to organize our work:
[tex]\[
\begin{array}{|l|l|l|}
\hline
x & x - \bar{x} & (x - \bar{x})^2 \\
\hline
28 & 28 - 33.0 = -5.0 & (-5.0)^2 = 25.0 \\
34 & 34 - 33.0 = 1.0 & (1.0)^2 = 1.0 \\
27 & 27 - 33.0 = -6.0 & (-6.0)^2 = 36.0 \\
42 & 42 - 33.0 = 9.0 & (9.0)^2 = 81.0 \\
52 & 52 - 33.0 = 19.0 & (19.0)^2 = 361.0 \\
15 & 15 - 33.0 = -18.0 & (-18.0)^2 = 324.0 \\
\hline
& & \text{Sum: } 828.0 \\
\hline
\end{array}
\][/tex]
So, the sum of the squares of the differences between each data value and the mean is [tex]\(828.0\)[/tex].
### (d) Standard deviation of the data set
The standard deviation provides a measure of the spread of the data values. Since we are dealing with a population (not a sample), we use the following formula for standard deviation [tex]\(\sigma\)[/tex]:
[tex]\[
\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{N}}
\][/tex]
Where [tex]\(N\)[/tex] is the number of data values (in this case, 6).
Given that the sum from Part (c) is [tex]\(828.0\)[/tex]:
[tex]\[
\sigma = \sqrt{\frac{828.0}{6}} = \sqrt{138.0} \approx 11.7473
\][/tex]
Therefore, the standard deviation of the data set is approximately [tex]\(11.7473\)[/tex].
