Question

finish times (to the nearest hour) for 10 dogsled teams are shown below.

make a frequency table showing class limits, class boundaries, midpoints, frequency, relative frequencies, and cumulative frequencies. use three classes. the class size of the given data is 24. (round your answer for relative frequency to the nearest hundredth and for midpoint to the nearest tenth.)

236 310 262 301 272 299 282 295 287 246

Explanation:

Step1: Find the minimum and maximum values

First, we identify the minimum and maximum values in the data set. The data points are 236, 310, 262, 301, 272, 299, 282, 295, 287, 246. The minimum value ($min$) is 236 and the maximum value ($max$) is 310.

Step2: Determine the class limits

We are told to use three classes with a class size of 24. We start the first class at the minimum value.
- First class: Lower class limit = 236, Upper class limit = 236 + 24 - 1 = 259 (since class size is 24, the upper limit is lower limit + class size - 1)
- Second class: Lower class limit = 260, Upper class limit = 260 + 24 - 1 = 283
- Third class: Lower class limit = 284, Upper class limit = 284 + 24 - 1 = 307 (Wait, but our maximum value is 310. Wait, maybe we made a mistake. Wait, the class size is 24, so let's re - calculate. Let's find the range: $310 - 236=74$. If we use three classes, the class width should be such that $74\div3\approx24.67$, but the problem says the class size is 24. So we adjust. Let's start the first class at 236.
First class: 236 - 259 (since 236+24 - 1 = 259)
Second class: 260 - 283 (260 + 24 - 1=283)
Third class: 284 - 307 (284+24 - 1 = 307), but we have a data point 310. Wait, maybe the problem has a typo or we misinterpret. Wait, the data points: 236, 246 (first class: 236 - 259: values 236, 246 → frequency 2)
Second class: 260 - 283: 262, 272, 282 → frequency 3
Third class: 284 - 307: 287, 295, 299, 301 → frequency 4? But 310 is left? Wait, maybe the class size is 24, and we should have three classes. Let's recalculate the class boundaries properly.

Class boundaries: For a class with lower limit $L$ and upper limit $U$, the lower boundary is $L - 0.5$ and upper boundary is $U+ 0.5$ (since the data is to the nearest hour, which is a discrete variable, but we treat it as continuous for boundaries).

Let's list the data in order: 236, 246, 262, 272, 282, 287, 295, 299, 301, 310

We need three classes. Let's find the class width. The range is $310 - 236 = 74$. If we use three classes, the class width $w=\lceil\frac{74}{3}
ceil = 25$, but the problem says class size is 24. Let's proceed with the problem's instruction of class size 24.

First class: Lower limit = 236, Upper limit = 236+24 - 1=259. So data points in this class: 236, 246 → frequency = 2
Second class: Lower limit = 260, Upper limit = 260 + 24-1 = 283. Data points: 262, 272, 282 → frequency = 3
Third class: Lower limit = 284, Upper limit = 284+24 - 1=307. Data points: 287, 295, 299, 301 → frequency = 4. But we have 310 left. Wait, maybe the upper limit of the third class is 310? No, the problem says class size is 24. Maybe there is an error in the problem, but we will proceed with the given data.

Step3: Calculate class boundaries

- First class: Lower boundary = $236 - 0.5=235.5$, Upper boundary = $259 + 0.5 = 259.5$
- Second class: Lower boundary = $260 - 0.5 = 259.5$, Upper boundary = $283+0.5 = 283.5$
- Third class: Lower boundary = $284 - 0.5=283.5$, Upper boundary = $307 + 0.5=307.5$ (But we have 310, which is above 307.5. Maybe the class size is 25? Let's check the range again. $310 - 236 = 74$. $74\div3\approx24.67$, so class size 25. Let's try that.

First class: 236 - 260 (236+25 - 1 = 260)
Data points: 236, 246 → frequency = 2
Second class: 261 - 285 (261+25 - 1 = 285)
Data points: 262, 272, 282, 287? No, 287 is above 285. Wait, 282 is in, 287 is out. So 262, 272, 282 → frequency = 3
Third class: 286 - 310 (286+25 - 1 = 310)
Data points: 287, 295, 299, 301, 310 → frequency = 5. But the total number of data points is 10. 2 + 3+5 = 10. Ah, this makes sense. Maybe the class size is 25 (since 74 divided by 3 is approximately 25). The problem might have a typo, but let's proceed with this as the sum of frequencies is 10.

Let's recast with class size 25 (since 3 classes, range 74, 74/3≈25)

Class Limits:

- Class 1: 236 - 260
- Class 2: 261 - 285
- Class 3: 286 - 310

Class Boundaries:

- Class 1: Lower boundary = 235.5, Upper boundary = 260.5
- Class 2: Lower boundary = 260.5, Upper boundary = 285.5
- Class 3: Lower boundary = 285.5, Upper boundary = 310.5

Midpoints:

Midpoint of a class is calculated as $\frac{\text{Lower class limit}+\text{Upper class limit}}{2}$

- Class 1: $\frac{236 + 260}{2}=\frac{496}{2}=248.0$
- Class 2: $\frac{261+285}{2}=\frac{546}{2}=273.0$
- Class 3: $\frac{286 + 310}{2}=\frac{596}{2}=298.0$

Frequency:

- Class 1: Number of data points between 236 and 260 (inclusive): 236, 246 → frequency ($f$) = 2
- Class 2: Number of data points between 261 and 285 (inclusive): 262, 272, 282 → frequency ($f$) = 3
- Class 3: Number of data points between 286 and 310 (inclusive): 287, 295, 299, 301, 310 → frequency ($f$) = 5

Relative Frequency:

Relative frequency ($rf$) is calculated as $\frac{f}{n}$, where $n = 10$ (total number of data points)

- Class 1: $rf=\frac{2}{10}=0.20$
- Class 2: $rf=\frac{3}{10}=0.30$
- Class 3: $rf=\frac{5}{10}=0.50$

Cumulative Frequency:

Cumulative frequency ($cf$) is the sum of frequencies up to the current class.

- Class 1: $cf = 2$
- Class 2: $cf=2 + 3=5$
- Class 3: $cf=5 + 5 = 10$

Now, let's create the frequency table:

| Class Limits | Class Boundaries | Midpoint | Frequency | Relative Frequency | Cumulative Frequency |
|--------------|------------------|----------|-----------|--------------------|----------------------|
| 236 - 260 | 235.5 - 260.5 | 248.0 | 2 | 0.20 | 2 |
| 261 - 285 | 260.5 - 285.5 | 273.0 | 3 | 0.30 | 5 |
| 286 - 310 | 285.5 - 310.5 | 298.0 | 5 | 0.50 | 10 |

If we strictly follow the class size of 24 as per the problem:

Class Limits:

- Class 1: 236 - 259
- Class 2: 260 - 283
- Class 3: 284 - 307

Class Boundaries:

- Class 1: 235.5 - 259.5
- Class 2: 259.5 - 283.5
- Class 3: 283.5 - 307.5

Midpoints:

- Class 1: $\frac{236 + 259}{2}=\frac{495}{2}=247.5$
- Class 2: $\frac{260+283}{2}=\frac{543}{2}=271.5$
- Class 3: $\frac{284 + 307}{2}=\frac{591}{2}=295.5$

Frequency:

- Class 1: 236, 246 → 2
- Class 2: 262, 272, 282 → 3
- Class 3: 287, 295, 299, 301 → 4 (we have 310 left, which is an issue as total frequency would be 2 + 3+4 = 9, missing one data point)

Since the problem says "the class size of the given data is 24", we will proceed with the class size 24, and note that there is a data point (310) that does not fit, which might be an error. But following the problem's instruction:

Frequency Table (Class size = 24):

| Class Limits | Class Boundaries | Midpoint | Frequency | Relative Frequency | Cumulative Frequency |
|--------------|------------------|----------|-----------|--------------------|----------------------|
| 236 - 259 | 235.5 - 259.5 | 247.5 | 2 | 0.20 | 2 |
| 260 - 283 | 259.5 - 283.5 | 271.5 | 3 | 0.30 | 5 |
| 284 - 307 | 283.5 - 307.5 | 295.5 | 4 | 0.40 | 9 |
| (Note: Data point 310 is unclassified) |

But since the total number of data points is 10, the correct approach is to have a class that includes 310. So the class size should be adjusted. The most probable class size is 25 (as 74/3≈25) to include all data points.

Answer:

The frequency table (with class size 25 to include all data points) is as follows:

Class Limits	Class Boundaries	Midpoint	Frequency	Relative Frequency	Cumulative Frequency
261 - 285	260.5 - 285.5	273.0	3	0.30	5
286 - 310	285.5 - 310.5	298.0	5	0.50	10

If we follow the class size of 24 (with a missing data point):

Class Limits	Class Boundaries	Midpoint	Frequency	Relative Frequency	Cumulative Frequency
260 - 283	259.5 - 283.5	271.5	3	0.30	5
284 - 307	283.5 - 307.5	295.5	4	0.40	9