Chapter 5.1, Problem 1AC

Dropping College Courses

Use the following table to answer the questions.

Reason for dropping a college course	Frequency	Percentage
Too difficult	458
Illness	40
Change in work schedule	20
Change of major	14
Family-related problems	9
Money	7
Miscellaneous	6
No meaningful reason	3

1. What is the variable under study? Is it a random variable?

2. How many people were in the study?

3. Complete the table.

4. From the information given, what is the probability that a student will drop a class because of illness? Money? Change of major?

5. Would you consider the information in the table to be a probability distribution?

6. Are the categories mutually exclusive?

7. Are the categories independent?

8. Are the categories exhaustive?

9. Are the two requirements for a discrete probability distribution met?

To determine

To identify: The variable under study.

To check: Whether the variable is random variable or not.

Answer to Problem 1AC

The variable under study is “Reason for dropping a college course”.

Yes, the variable is random variable.

Explanation of Solution

Given info:

The table shows the frequency for the dropping a college course.

Justification:

Random variable:

If the values of variable are drawn randomly then the variable is said to be random variable.

From the given information the variable is “Reason for dropping a college course”.

Here, the values in the variable “Reason for dropping a college course” are drawn randomly. Thus, the variable “Reason for dropping a college course” is random variable.

To determine

To identify: The number of people in the study.

Answer to Problem 1AC

The number of people in the study is 144.

Explanation of Solution

Calculation:

The total number of people in the study is 144 $\left(=45+40+20+14+9+7+6+3\right)$.

To determine

To complete: The table.

Answer to Problem 1AC

The complete table is,

Reason for dropping a college course	Frequency	Percentage
Too difficult	45	31.25
Illness	40	27.78
Change in work schedule	20	13.89
Change of major	14	9.72
Family-related problems	9	6.25
Money	7	4.86
Miscellaneous	6	4.17
No meaningful reason	3	2.08
Total	144

Explanation of Solution

Calculation:

The percentage of the frequency is calculated as given in the table follows,

Reason for dropping a college course	Frequency	Percentage
Too difficult	45	$\frac{45}{144}\times 100=31.25$
Illness	40	$\frac{40}{144}\times 100=27.78$
Change in work schedule	20	$\frac{20}{144}\times 100=13.89$
Change of major	14	$\frac{14}{144}\times 100=9.72$
Family-related problems	9	$\frac{9}{144}\times 100=6.25$
Money	7	$\frac{7}{144}\times 100=4.86$
Miscellaneous	6	$\frac{6}{144}\times 100=4.17$
No meaningful reason	3	$\frac{3}{144}\times 100=2.08$
Total	144

To determine

To find: The probabilities that the student will drop a class because of illness, money and change of major.

Answer to Problem 1AC

The probabilities that the student will drop a class because of illness, money and change of major are 0.28, 0.05 and 0.10.

Explanation of Solution

Calculation:

The numbers of students drop for illness, money and change of major are is 40, 7 and 14.

The total frequency is 144.

The probabilities that the student will drop a class because of illness is,

$\begin{array}{c}P\left(\text{Illness}\right)=\frac{40}{144}\\ =0.28\end{array}$

The probabilities that the student will drop a class because of money is,

$\begin{array}{c}P\left(\text{Money}\right)=\frac{7}{144}\\ =0.05\end{array}$

The probabilities that the student will drop a class because of money is,

$\begin{array}{c}P\left(\text{Change of major}\right)=\frac{14}{144}\\ =0.10\end{array}$

Thus, the probabilities that the student will drop a class because of illness, money and change of major are 0.28, 0.05 and 0.10.

To determine

To check: Whether the information in the table consider as to be a probability distribution.

Answer to Problem 1AC

The information in the table is not itself considered as a probability distribution. But it can used as one.

Explanation of Solution

Justification:

Here, the information in the table is not itself considered as a probability distribution. But the information can be used as one.

To determine

To check: Whether the categories are mutually exclusive.

Answer to Problem 1AC

No, the categories are not mutually exclusive.

Explanation of Solution

Justification:

Here, the categories are used for finding the probability but these categories are not mutually exclusive.

To determine

To check: Whether the categories are independent.

Answer to Problem 1AC

No, the categories are not independent.

Explanation of Solution

Justification:

Here, the categories are not independent because the categories depend on the other variables.

To determine

To check: Whether the categories are exhaustive.

Answer to Problem 1AC

Yes, the categories are exhaustive.

Explanation of Solution

Justification:

Here, the categories are exhaustive because it occur compulsory.

To determine

To check: Whether the requirement for a discrete probability distribution met or not.

Answer to Problem 1AC

Yes, the requirement for a discrete probability distribution met.

Explanation of Solution

Justification:

Requirements for a Probability Distribution:

• The sum of probabilities of all events in the sample space must be equal to 1. That is, ${\displaystyle \sum P\left(X\right)}=1$
• The probability values of each event in the sample space must be between or equal to 0 and 1. That is, $0\le P\left(X\right)\le 1$

From the part of (3), it can be observed that all the probabilities are between 0 and 1. Also the sum of the probabilities is equal to 1.

Thus, the distribution met the requirements of discrete probability distribution.