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Chapter 7: Problem 9

The grade distribution for a certain class is shown in the following table. Find the probability distribution associated with these data. $$ \begin{array}{lllll} \hline \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\ \hline \text { Frequency of } & & & & & \\ \text { Occurrence } & 4 & 10 & 18 & 6 & 2 \\ \hline \end{array} $$

Short Answer

Expert verified
The probability distribution associated with the grade distribution data is: $$ \begin{array}{lllll} \hline \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\ \text { Probability } & \frac{4}{40} & \frac{10}{40} & \frac{18}{40} & \frac{6}{40} & \frac{2}{40} \\ \hline \end{array} $$

Step by step solution

01

Determine the total number of students

We can find the total number of students in the class by adding the frequencies of each grade. We are given the following data: - A: 4 students - B: 10 students - C: 18 students - D: 6 students - F: 2 students Total number of students: \(4 + 10 + 18 + 6 + 2 = 40\) There are a total of 40 students in the class.
02

Calculate the probability for each grade

Now, we will find the probability (P) of obtaining each grade (A, B, C, D, and F), by dividing the frequency of occurrences of the grade by the total number of students in the class. 1. P(A): Probability of getting grade A is \(\frac{\text{Number of students with grade A}}{\text{Total number of students}} = \frac{4}{40}\) 2. P(B): Probability of getting grade B is \(\frac{\text{Number of students with grade B}}{\text{Total number of students}} = \frac{10}{40}\) 3. P(C): Probability of getting grade C is \(\frac{\text{Number of students with grade C}}{\text{Total number of students}} = \frac{18}{40}\) 4. P(D): Probability of getting grade D is \(\frac{\text{Number of students with grade D}}{\text{Total number of students}} = \frac{6}{40}\) 5. P(F): Probability of getting grade F is \(\frac{\text{Number of students with grade F}}{\text{Total number of students}} = \frac{2}{40}\)
03

Present the probability distribution

Now we have calculated the probability for each grade, we can present the probability distribution associated with the grade distribution data. $$ \begin{array}{lllll} \hline \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\ \text { Probability } & \frac{4}{40} & \frac{10}{40} & \frac{18}{40} & \frac{6}{40} & \frac{2}{40} \\ \hline \end{array} $$ Therefore, the probability distribution associated with the given data is as shown in the table above.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grade Distribution
Grade distribution refers to how the grades of a group of students are spread across different categories. Digitally, you can think of it like a visual representation that shows how many students received each grade, such as A, B, C, D, or F. Understanding the grade distribution in a class helps educators gain insight into the overall performance of students. It can also highlight which areas may need more attention or where students excel. For example, if most students receive a C, this may suggest a need for better instruction or support in that subject.
Frequency Distribution
Frequency distribution is a crucial concept in statistics that expresses how often each value in a set of data occurs. In the context of our exercise, it indicates how many students received each grade. This is represented by associating each grade with its respective frequency:
  • Grade A: 4 students
  • Grade B: 10 students
  • Grade C: 18 students
  • Grade D: 6 students
  • Grade F: 2 students
A better grasp of frequency distribution can help educators see common patterns or trends in student grades, which is essential for curriculum improvement.
Total Frequency Calculation
Total frequency calculation involves summing up all individual frequencies to find the overall dataset size. In our exercise, this means finding the total number of students. Here's how we calculated it:
  • We add the number of students for each grade: 4 (A) + 10 (B) + 18 (C) + 6 (D) + 2 (F).
  • This gives us a total of 40 students.
Knowing the total frequency is vital when calculating probabilities since it serves as the denominator for each probability calculation.
Probability Calculation
Probability calculation is all about determining the likelihood of a specific event based on available data. In this exercise, we calculate the probability of each grade by dividing the frequency of that grade by the total number of students.For instance, to find the probability of a student receiving an A:
  • Use the formula: \( P(A) = \frac{\text{Number of A's}}{\text{Total students}} \)
  • Substitute the numbers: \( P(A) = \frac{4}{40} \)
Using this method, we find the probability for each grade:
  • \( P(B) = \frac{10}{40} \)
  • \( P(C) = \frac{18}{40} \)
  • \( P(D) = \frac{6}{40} \)
  • \( P(F) = \frac{2}{40} \)
The collective results form what is called a probability distribution. It illustrates how likely it is to randomly select a student with a specific grade from the class.

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Most popular questions from this chapter

In a past presidential election, it was estimated that the probability that the Republican candidate would be elected was \(\frac{3}{5}\), and therefore the probability that the Democratic candidate would be elected was \(\frac{2}{5}\) (the two Independent candidates were given no chance of being elected). It was also estimated that if the Republican candidate were elected, the probability that a conservative, moderate, or liberal judge would be appointed to the Supreme Court (one retirement was expected during the presidential term) was \(\frac{1}{2}, \frac{1}{3}\), and \(\frac{1}{6}\), respectively. If the Democratic candidate were elected, the probabilities that a conservative, moderate, or liberal judge would be appointed to the Supreme Court would be \(\frac{1}{8}, \frac{3}{8}\), and \(\frac{1}{2}\), respectively. A conservative judge was appointed to the Supreme Court during the presidential term. What is the probability that the Democratic candidate was elected?

There were 80 male guests at a party. The number of men in each of four age categories is given in the following table. The table also gives the probability that a man in the respective age category will keep his paper money in order of denomination. $$\begin{array}{lcc} \hline \text { Age } & \text { Men } & \text { Keep Paper Money in Order, \% } \\\ \hline 21-34 & 25 & 9 \\ \hline 35-44 & 30 & 61 \\ \hline 45-54 & 15 & 80 \\ \hline 55 \text { and over } & 10 & 80 \\ \hline \end{array}$$ A man's wallet was retrieved and the paper money in it was kept in order of denomination. What is the probability that the wallet belonged to a male guest between the ages of 35 and 44 ?

The accompanying data were obtained from the financial aid office of a certain university: \begin{tabular}{lccc} \hline & \multicolumn{3}{c} { Not } & \\ & Receiving Financial Aid & Receiving Financial Aid & Total \\ \hline Undergraduates & \(4.222\) & 3,898 & 8,120 \\ \hline Graduates & \(1.879\) & 731 & 2,610 \\ \hline Total & \(6.101\) & 4,629 & 10,730 \\ \hline \end{tabular} Let \(A\) be the event that a student selected at random from this university is an undergraduate student, and let \(B\) be the event that a student selected at random is receiving financial aid. a. Find each of the following probabilities: \(P(A), P(B)\), \(P(A \cap B), P(B \mid A)\), and \(P\left(B \mid A^{\circ}\right)\) b. Are the events \(A\) and \(B\) independent events?

Data compiled by the Highway Patrol Department regarding the use of seat belts by drivers in a certain area after the passage of a compulsory seat-belt law are shown in the accompanying table. $$\begin{array}{lcc} \hline & & \text { Percent of } \\ \text { Drivers } & \begin{array}{c} \text { Percent } \\ \text { of Drivers } \\ \text { in Group } \end{array} & \begin{array}{c} \text { Group Stopped } \\ \text { for Moving } \end{array} \\ \hline \text { Group I } & \text { Violation } \\ \text { (using seat belts) } & 64 & .2 \\ \hline \text { Group II } & & \\ \text { (not using seat belts) } & 36 & .5 \\ \hline \end{array}$$ If a driver in that area is stopped for a moving violation, what is the probability that he or she a. Will have a seat belt on? b. Will not have a seat belt on?

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$\begin{array}{llllll} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array}$$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)
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