cmariacardona2608
Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.