Question

question 7 of 9
stacked
the histograms applet displays the travel times to campus in minutes on a particular tuesday for a sample of 20 students.
now set the applet to 8 classes.
in which class do most of the data values lie? (click the points on the x - axis to see the class boundaries.)
lower bound of the class:
upper bound of the class:

Explanation:

Step1: Determine the range of data

The travel times range from 0 to 40 minutes (since the x - axis goes from 0 to 40). We need to divide this range into 8 classes. The class width \(w=\frac{40 - 0}{8}=5\). So the classes will be: \(0 - 5\), \(5 - 10\), \(10 - 15\), \(15 - 20\), \(20 - 25\), \(25 - 30\), \(30 - 35\), \(35 - 40\).

Step2: Analyze the original histogram (with 6 classes)

In the original histogram (6 classes), the tallest bar is around the middle. When we change to 8 classes (class width 5), we need to see which class has the most data. Looking at the original data distribution, the peak is around 10 - 20 in the 6 - class histogram. When we split into 8 classes (width 5), the class \(10 - 15\) or \(15 - 20\)? Wait, no, let's recalculate. Wait, the original x - axis has points at 0, 10, 20, 30. Wait, maybe the initial range is from 5 to 30? Wait, no, the first bar starts at 5 (since the first point is at 5? Wait, the first green dot is at 5? Wait, looking at the graph, the first bar's left end is at 5? Wait, the x - axis has marks at 0, 10, 20, 30, 40. The first bar is from 5 to 10? Wait, maybe I made a mistake. Let's look at the count. The first bar (leftmost) has a count of 3, then 2, then 7, then 4, then 2, then 2. Wait, when we set to 8 classes, the class width is \(\frac{40}{8}=5\). So classes are \(0 - 5\), \(5 - 10\), \(10 - 15\), \(15 - 20\), \(20 - 25\), \(25 - 30\), \(30 - 35\), \(35 - 40\). Now, looking at the original data, the tallest bar in the 6 - class histogram is between 10 and 20 (since the third bar is the tallest, with count 7). When we split into 8 classes, the class \(10 - 15\) and \(15 - 20\) are within 10 - 20. But let's check the original bar positions. The first bar is from 5 to 10 (count 3), second from 10 to 15 (count 2), third from 15 to 20 (count 7), fourth from 20 to 25 (count 4), fifth from 25 to 30 (count 2), sixth from 30 to 35 (count 2). Wait, no, maybe the initial class boundaries are different. Wait, the problem says "Click the points on the x - axis to see the class boundaries". But since we can infer, when we set to 8 classes, the class width is 5. The most data (from the original histogram, the tallest bar is at 15 - 20? Wait, no, the third bar (after 5 - 10, 10 - 15) is 15 - 20 with count 7. Wait, no, the first bar is 5 - 10 (count 3), second 10 - 15 (count 2), third 15 - 20 (count 7), fourth 20 - 25 (count 4), fifth 25 - 30 (count 2), sixth 30 - 35 (count 2). Wait, no, maybe the x - axis points are at 5, 10, 15, 20, 25, 30, 35, 40. Wait, the original graph has a first bar starting at 5 (the first green dot is at 5). So when we set to 8 classes, the classes are \(0 - 5\), \(5 - 10\), \(10 - 15\), \(15 - 20\), \(20 - 25\), \(25 - 30\), \(30 - 35\), \(35 - 40\). But in the original histogram (6 classes), the tallest bar is at 15 - 20 (count 7). So when we split into 8 classes, the class \(15 - 20\) (lower bound 15, upper bound 20) or wait, no, the class with the most data. Wait, maybe I messed up the class width. Wait, the range is from 0 to 40, 8 classes, so class width is 5. So the classes are:

1. \(0 - 5\)
2. \(5 - 10\)
3. \(10 - 15\)
4. \(15 - 20\)
5. \(20 - 25\)
6. \(25 - 30\)
7. \(30 - 35\)
8. \(35 - 40\)

Looking at the original histogram (6 classes), the first bar (count 3) is \(5 - 10\), second (count 2) \(10 - 15\), third (count 7) \(15 - 20\), fourth (count 4) \(20 - 25\), fifth (count 2) \(25 - 30\), sixth (count 2) \(30 - 35\). So when we go to 8 classes, the class \(15 - 20\) (lower bound 15, upper bound 20) will still have a high count. Wait, but the question is to find the class with most data when set to 8 classes. So the lower bound of the class is 10? No, wait, no. Wait, maybe the initial range is from 0 to 40, and when we set to 8 classes, the class with the most data is \(10 - 15\)? No, let's think again. Wait, the original histogram has a bar with count 7 between 10 and 20 (in 6 - class). When we split into 8 classes (width 5), the class \(10 - 15\) and \(15 - 20\) are within that. But the tallest bar in the original is at 15 - 20 (count 7). So when we set to 8 classes, the class with the most data is \(15 - 20\)? Wait, no, maybe the lower bound is 10? Wait, I think I made a mistake. Let's check the class width again. If the total range is 40 (from 0 to 40), 8 classes, so class width is 5. So the first class is 0 - 5, second 5 - 10, third 10 - 15, fourth 15 - 20, fifth 20 - 25, sixth 25 - 30, seventh 30 - 35, eighth 35 - 40. Now, looking at the original data (6 classes), the counts are: 3 (5 - 10), 2 (10 - 15), 7 (15 - 20), 4 (20 - 25), 2 (25 - 30), 2 (30 - 35). So when we split into 8 classes, the class 15 - 20 (count 7 in 6 - class, which will be split into 15 - 20 (count still high) and 20 - 25 (count 4). Wait, no, the 15 - 20 class in 6 - class is one class, in 8 - class it's two classes: 15 - 20 and 20 - 25? No, no, 15 - 20 is one class (width 5) in 8 - class. Wait, no, 15 - 20 is a class with width 5. So the count in 15 - 20 (8 - class) will be the same as the count in 15 - 20 (6 - class) if the data is within that range. Wait, the original 6 - class has a class 10 - 20 (width 10) with count 2 + 7=9? No, no, the original 6 - class: first bar 5 - 10 (count 3), second 10 - 15 (count 2), third 15 - 20 (count 7), fourth 20 - 25 (count 4), fifth 25 - 30 (count 2), sixth 30 - 35 (count 2). So total data points: 3+2 + 7+4+2+2 = 20, which matches the sample size. Now, when we split into 8 classes (width 5), the classes are:

- 0 - 5: count 0 (since original data starts at 5)
- 5 - 10: count 3 (same as original first bar)
- 10 - 15: count 2 (same as original second bar)
- 15 - 20: count 7 (same as original third bar)
- 20 - 25: count 4 (same as original fourth bar)
- 25 - 30: count 2 (same as original fifth bar)
- 30 - 35: count 2 (same as original sixth bar)
- 35 - 40: count 0

Wait, that can't be, because we divided into 8 classes but the data is from 5 - 35. Wait, maybe the range is from 5 to 40? Then class width is \(\frac{40 - 5}{8}\approx4.375\), but usually we take integer class width. Maybe the x - axis is from 0 to 40, and the data is from 5 to 35. But the problem says "set the applet to 8 classes". So the applet will automatically calculate the class boundaries. Looking at the original graph, the first green dot is at 5, then 10, 15, 20, 25, 30, 35, 40? Wait, the x - axis has points at 0, 10, 20, 30, but when we set to 8 classes, the applet will create classes with width \(\frac{40}{8}=5\), so starting at 0. But the original data has no points below 5 (since the first bar starts at 5). So the class with the most data is 15 - 20 (lower bound 15, upper bound 20)? No, wait, the original third bar (count 7) is from 15 - 20, so when we set to 8 classes, the class with the highest count is 15 - 20, so the lower bound is 10? No, I'm confused. Wait, maybe the correct class is 10 - 15? No, let's look at the initial histogram again. The first bar (left) has count 3, second count 2, third count 7 (tallest), fourth count 4, fifth count 2, sixth count 2. So the tallest bar is the third one, which is between 10 and 20 (in 6 - class). When we split into 8 classes (width 5), the third bar (10 - 20 in 6 - class) will be split into two classes: 10 - 15 and 15 - 20. The count in 10 - 15 is 2 (from original second bar) plus part of the third bar? No, no, the original second bar is 10 - 15 (count 2), third bar is 15 - 20 (count 7). So in 8 - class, 10 - 15 (count 2), 15 - 20 (count 7), 20 - 25 (count 4), etc. So the tallest class is 15 - 20, so lower bound is 15, upper bound is 20? But that doesn't seem right. Wait, maybe the class width is 5, and the first class is 0 - 5, but there's no data there. The first class with data is 5 - 10 (count 3), then 10 - 15 (count 2), then 15 - 20 (count 7), then 20 - 25 (count 4), then 25 - 30 (count 2), then 30 - 35 (count 2), then 35 - 40 (count 0). So the class with the most data is 15 - 20, so lower bound is 15, upper bound is 20. Wait, but maybe I made a mistake. Alternatively, maybe the class is 10 - 15? No, the count in 10 - 15 is 2, in 15 - 20 is 7. So the lower bound is 15, upper bound is 20.

Answer:

lower bound of the class: \(15\)

upper bound of the class: \(20\)