Answered

Thirty-two people were chosen at random from employees of a large company. Their commute times (in hours) were recorded in a table shown below.

\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline 0.5 & 0.9 & 0.5 & 0.4 & 0.6 & 1.3 & 1.1 & 0.6 \\
\hline 0.7 & 0.4 & 0.8 & 1.1 & 0.9 & 1.5 & 0.4 & 1.0 \\
\hline 0.9 & 1.0 & 0.6 & 0.2 & 0.7 & 1.1 & 0.6 & 1.2 \\
\hline 0.4 & 1.2 & 0.6 & 0.7 & 1.3 & 0.8 & 0.4 & 0.9 \\
\hline
\end{tabular}

Construct a frequency table using a class interval width of 0.2 starting at 0.15.

\begin{tabular}{|c|c|c|}
\hline Class Interval & Frequency & Relative Frequency \\
\hline [tex]$0.15-0.35$[/tex] & 2 & [tex]$\frac{2}{32}$[/tex] \\
\hline [tex]$0.35-0.55$[/tex] & 7 & [tex]$\frac{7}{32}$[/tex] \\
\hline [tex]$0.55-0.75$[/tex] & 8 & [tex]$\frac{8}{32}$[/tex] \\
\hline [tex]$0.75-0.95$[/tex] & 7 & [tex]$\frac{7}{32}$[/tex] \\
\hline [tex]$0.95-1.15$[/tex] & 6 & [tex]$\frac{6}{32}$[/tex] \\
\hline [tex]$1.15-1.35$[/tex] & 2 & [tex]$\frac{2}{32}$[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Let's work through the problem step-by-step to construct the frequency table using the given class intervals.

Given Data:

List of commute times (in hours):
[tex]\[ [0.5, 0.9, 0.5, 0.4, 0.6, 1.3, 1.1, 0.6, 0.7, 0.4, 0.8, 1.1, 0.9, 1.5, 0.4, 1.0, 0.9, 1.0, 0.6, 0.2, 0.7, 1.1, 0.6, 1.2, 0.4, 1.2, 0.6, 0.7, 1.3, 0.8, 0.4, 0.9] \][/tex]

Class Intervals (width = 0.2, starting at 0.15):
[tex]\[ [(0.15, 0.35), (0.35, 0.55), (0.55, 0.75), (0.75, 0.95), (0.95, 1.15), (1.15, 1.35), (1.35, 1.55)] \][/tex]

Counts (Frequency) and Relative Frequency calculations are as follows:

### Frequency Calculation:
Count how many data points fall in each interval:

- [tex]\(0.15 - 0.35\)[/tex]
- 0.2, 0.4
- Frequency: 1

- [tex]\(0.35 - 0.55\)[/tex]
- 0.5, 0.5, 0.4, 0.4, 0.4
- Frequency: 7

- [tex]\(0.55 - 0.75\)[/tex]
- 0.6, 0.6, 0.7, 0.6, 0.6, 0.6, 0.7, 0.7
- Frequency: 8

- [tex]\(0.75 - 0.95\)[/tex]
- 0.8, 0.9, 0.9, 0.8, 0.9, 0.8
- Frequency: 6

- [tex]\(0.95 - 1.15\)[/tex]
- 1.0, 1.0, 1.1, 1.1, 1.0
- Frequency: 5

- [tex]\(1.15 - 1.35\)[/tex]
- 1.2, 1.2, 1.3, 1.2, 1.3
- Frequency: 4

- [tex]\(1.35 - 1.55\)[/tex]
- 1.5
- Frequency: 1

### Relative Frequency Calculation:
Relative frequency is calculated as the frequency of each interval divided by the total number of data points (32).

- [tex]\(0.15 - 0.35\)[/tex]: [tex]\(\frac{1}{32} = 0.03125\)[/tex]
- [tex]\(0.35 - 0.55\)[/tex]: [tex]\(\frac{7}{32} \approx 0.21875\)[/tex]
- [tex]\(0.55 - 0.75\)[/tex]: [tex]\(\frac{8}{32} = 0.25\)[/tex]
- [tex]\(0.75 - 0.95\)[/tex]: [tex]\(\frac{6}{32} = 0.1875\)[/tex]
- [tex]\(0.95 - 1.15\)[/tex]: [tex]\(\frac{5}{32} = 0.15625\)[/tex]
- [tex]\(1.15 - 1.35\)[/tex]: [tex]\(\frac{4}{32} = 0.125\)[/tex]
- [tex]\(1.35 - 1.55\)[/tex]: [tex]\(\frac{1}{32} = 0.03125\)[/tex]

### Final Frequency Table:

[tex]\[ \begin{tabular}{|c|c|c|} \hline Class Interval & Frequency & Relative Frequency \\ \hline $0.15-0.35$ & 1 & $0.03125$ \\ $0.35-0.55$ & 7 & $0.21875$ \\ $0.55-0.75$ & 8 & $0.25$ \\ $0.75-0.95$ & 6 & $0.1875$ \\ $0.95-1.15$ & 5 & $0.15625$ \\ $1.15-1.35$ & 4 & $0.125$ \\ $1.35-1.55$ & 1 & $0.03125$ \\ \hline \end{tabular} \][/tex]

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