Question

the accompanying data are 45 commute times to work in minutes for workers of age 16 or older in chicago. construct a frequency distribution. use a class width of 15 minutes and begin with a lower class limit of 0 minutes. do the data amounts appear to have a normal distribution? examine the data and identify anything appearing to be unique. click the icon to view the commute times. construct the frequency distribution. commute time (minutes) frequency 0–14 15–29 30–44 45–59 60–74 75–89 (type whole numbers.)

Answer:

To solve this, we first need the actual commute time data (which is referenced as "Click the icon to view the commute times" but not provided here). However, the general process to construct the frequency distribution is as follows:

Step 1: Define the classes

The class width is 15, starting at 0. So the classes are:

• \( 0 - 14 \) (since \( 0 + 15 - 1 = 14 \))
• \( 15 - 29 \) ( \( 15 + 15 - 1 = 29 \))
• \( 30 - 44 \) ( \( 30 + 15 - 1 = 44 \))
• \( 45 - 59 \) ( \( 45 + 15 - 1 = 59 \))
• \( 60 - 74 \) ( \( 60 + 15 - 1 = 74 \))
• \( 75 - 89 \) ( \( 75 + 15 - 1 = 89 \))

Step 2: Count frequencies

For each class, count how many commute times fall within that interval. For example, if the data is (hypothetical example): 5, 10, 12, 18, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 (and more to reach 45 data points), we would:

• For \( 0 - 14 \): count data points between 0 and 14 (inclusive). Suppose 3 data points: frequency = 3.
• For \( 15 - 29 \): count data points between 15 and 29 (inclusive). Suppose 5 data points: frequency = 5.
• For \( 30 - 44 \): count data points between 30 and 44 (inclusive). Suppose 8 data points: frequency = 8.
• For \( 45 - 59 \): count data points between 45 and 59 (inclusive). Suppose 10 data points: frequency = 10.
• For \( 60 - 74 \): count data points between 60 and 74 (inclusive). Suppose 9 data points: frequency = 9.
• For \( 75 - 89 \): count data points between 75 and 89 (inclusive). Suppose 10 data points: frequency = 10.

(Note: These are hypothetical frequencies. The actual frequencies depend on the given data.)

Step 3: Check for normal distribution

A normal distribution should have frequencies that increase to a peak and then decrease symmetrically. If the frequencies are roughly symmetric around a central class, it may be normal. Also, check for any unique features like outliers (data points far from others) or gaps (no data in a class).

Since the actual data is not provided, we can't give the exact frequencies, but the process is as above.

To complete the frequency distribution, you need to:

1. Obtain the 45 commute time data points.
2. For each class interval, count how many data points lie within it.
3. Fill in the frequency column with those counts.

For example, if the data was (a small subset for illustration):

Commute times: 5, 10, 12, 18, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 2, 8, 15, 19, 22, 28, 32, 38, 42, 46, 52, 58, 62, 68, 72, 78, 82, 1, 9, 16, 21, 27, 31, 37, 41, 47, 53

Then:

• \( 0 - 14 \): 5, 10, 12, 2, 8, 1, 9 → 7 data points (frequency = 7)
• \( 15 - 29 \): 18, 20, 25, 15, 19, 22, 28, 16, 21, 27 → 10 data points (frequency = 10)
• \( 30 - 44 \): 30, 35, 40, 32, 38, 42, 31, 37, 41 → 9 data points (frequency = 9)
• \( 45 - 59 \): 45, 50, 55, 46, 52, 58, 47, 53 → 8 data points (frequency = 8)
• \( 60 - 74 \): 60, 65, 70, 62, 68, 72 → 6 data points (frequency = 6)
• \( 75 - 89 \): 75, 80, 85, 78, 82 → 5 data points (frequency = 5)

(This is just an example. The actual frequencies will depend on the given data.)

After constructing the frequency distribution, to check for normal distribution, we look at the shape of the frequency distribution. A normal distribution should have a bell - shaped curve, meaning that the frequencies increase to a maximum (the peak) and then decrease in a roughly symmetric way around the peak. For example, if the frequencies for the classes \( 0 - 14 \), \( 15 - 29 \), \( 30 - 44 \), \( 45 - 59 \), \( 60 - 74 \), \( 75 - 89 \) are 7, 10, 12, 10, 5, 1 (hypothetical), this has a peak at \( 30 - 44 \) and is somewhat symmetric around it, suggesting a normal - like distribution. If there is a class with a very low frequency (an outlier class) or a large gap between frequencies, that would be a unique feature.

Since we don't have the actual data, we can't provide the exact frequency distribution, but the above steps show how to construct it and analyze for normal distribution and unique features.