A statistics professor asked students in a class their ages. On the basis of this information the professor states that the average age of all the students in the university is 27 years. This is an example of

19) Step 1: Identify the class intervals from Exhibit 3-1 (from the previous question). The class intervals are 0-9, 10-19, 20-29, 30-39.

Step 2: Calculate the class width for any interval. For a class interval like $a-b$, the class width is $b - a + 1$. Using the first class (0-9): Class width = $9 - 0 + 1 = 10$. Using the second class (10-19): Class width = $19 - 10 + 1 = 10$. The class width is consistent.

$\text{b. is 10}$

20) Step 1: Identify the classes that represent students working 19 hours or less from Exhibit 3-1. These are the classes 0-9 and 10-19.

Step 2: Sum the frequencies for these classes. Frequency for 0-9 hours = 20 students. Frequency for 10-19 hours = 50 students. Total number of students working 19 hours or less = $20 + 50 = 70$.

$\text{d. None of the above}$ (Since 70 is not an option, and the options are 40, 50, 90, None of the above, the correct choice is d. None of the above.)

21) Step 1: Identify the class that represents students working 9 hours or less from Exhibit 3-1. This is the class 0-9 hours.

Step 2: Identify the frequency for this class and the total frequency. Frequency for 0-9 hours = 20 students. Total frequency (N) = 200 students.

Step 3: Calculate the relative frequency. Relative frequency = $\frac{Frequency of class}{Total frequency}$ Relative frequency = $\frac{20}{200} = 0.1$.

The calculated relative frequency is 0.1, which is