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The data set shows the January 1 noon temperatures in degrees Fahrenheit for a particular city in each of the past 6 years:

[tex]\[
\begin{array}{llllll}
28 & 34 & 27 & 42 & 52 & 15
\end{array}
\][/tex]

(a) What is the five-number summary of the data set?

(b) What is the mean, [tex]\(\bar{x}\)[/tex], of the data set?

(c) What is the sum of the squares of the differences between each data value and the mean? Use the table to organize your work.

[tex]\[
\begin{tabular}{|l|l|l|}
\hline
$x$ & $x-\bar{x}$ & $(x-\bar{x})^2$ \\
\hline
28 & & \\
\hline
34 & & \\
\hline
27 & & \\
\hline
42 & & \\
\hline
52 & & \\
\hline
15 & & \\
\hline
& & Sum: \\
\hline
\end{tabular}
\][/tex]

(d) What is the standard deviation of the data set? Use the sum from Part (c) and show your work.

Sagot :

GinnyAnswer
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].
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