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Chapter 6: Problem 4

Grade distribution An instructor always assigns final grades such that \(20 \%\) are \(A, 40 \%\) are \(B, 30 \%\) are \(C,\) and \(10 \%\) are \(\mathrm{D} .\) The grade point scores are 4 for \(\mathrm{A}, 3\) for \(\mathrm{B}, 2\) for \(\mathrm{C},\) and 1 for \(\mathrm{D}\). a. Specify the probability distribution for the grade point score of a randomly selected student of this instructor. b. Find the mean of this probability distribution. Interpret it.

Short Answer

Expert verified
a. Probabilities: A = 0.2, B = 0.4, C = 0.3, D = 0.1. b. Mean = 2.7, indicating an average between B and C.

Step by step solution

01

Define the Probability Distribution

To calculate the probability distribution, we note that the probability of getting different grades must sum up to 1. Given percentages can be directly converted to probabilities. Thus, the probabilities assigned to each grade are:- Probability of receiving A, \( P(A) = 0.2 \)- Probability of receiving B, \( P(B) = 0.4 \)- Probability of receiving C, \( P(C) = 0.3 \)- Probability of receiving D, \( P(D) = 0.1 \)
02

Assign Point Scores

Each grade has an associated grade point score: - Score for grade A is 4 - Score for grade B is 3 - Score for grade C is 2 - Score for grade D is 1 These scores will be used to find the mean of the probability distribution.
03

Calculate the Mean of the Distribution

The mean of a probability distribution, often known as the expected value, is calculated by the formula:\[ \mu = \sum (x_i \cdot P(x_i)) \]Where \( x_i \) is the score and \( P(x_i) \) is the probability. Therefore:- Mean = \( 4 \times 0.2 + 3 \times 0.4 + 2 \times 0.3 + 1 \times 0.1 \)- Calculating each term, \( 4 \times 0.2 = 0.8 \), \( 3 \times 0.4 = 1.2 \), \( 2 \times 0.3 = 0.6 \), \( 1 \times 0.1 = 0.1 \)- Add these values: \( 0.8 + 1.2 + 0.6 + 0.1 = 2.7 \)Therefore, the mean grade point score is 2.7.
04

Interpret the Mean

The mean grade point score of 2.7 implies that, on average, a student in this class is likely to have a grade between a B and a C.

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Key Concepts

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

Mean of Probability Distribution
The mean of a probability distribution is a measure that provides us with an average outcome when we consider all possible scenarios, weighted by their probabilities. In simpler terms, it tells us what outcome we can expect most on average from our system. You might also hear it referred to as the "expected value." This gives us a central point or a balance point in the distribution where most probabilities seem to cluster. In our example regarding students' grades:
  • The probability distribution shows us how grades are distributed, with probabilities of 0.2 for an A, 0.4 for a B, 0.3 for a C, and 0.1 for a D.
  • The mean or expected value of this distribution was calculated to be 2.7.
This mean tells us that the typical grade point score, averaged over many students, falls between a B and a C. It's a useful measure for anticipating the general performance of students in this course.
Expected Value
The expected value is a central concept in probability and statistics, representing the weighted average of all possible values. Essentially, the expected value provides a single summary figure describing the entire distribution.
To determine the expected value, you multiply each possible outcome by its respective probability and then sum all these values. In our grade distribution case, for each grade, we have:
  • Grade A: Score = 4; Probability = 0.2; Contribution to expected value = 4 脳 0.2 = 0.8
  • Grade B: Score = 3; Probability = 0.4; Contribution to expected value = 3 脳 0.4 = 1.2
  • Grade C: Score = 2; Probability = 0.3; Contribution to expected value = 2 脳 0.3 = 0.6
  • Grade D: Score = 1; Probability = 0.1; Contribution to expected value = 1 脳 0.1 = 0.1
Summing these values gives us an expected grade point score of 2.7. This quantifies the average grade a student might expect to receive if grades were largely determined by historical probability.
Grade Point Score
A Grade Point Score (GPS) is a numerical representation of a student's performance, often used in educational settings to indicate a student's grade. Each letter grade is typically assigned a specific point value, making it easy to compute averages and overall academic performance.
  • In our scenario, we consider these point allocations: A = 4, B = 3, C = 2, D = 1.
  • The meaning behind these scores is straightforward: higher scores signal better grades.
  • When an instructor assigns constant numeric values to these letter grades, it allows for simple and efficient analysis of student performance over time.
Grasping how these scores transform into averages helps in understanding both individual performance and overall class achievement. As we've seen, through the mean and expected value calculations, the grade point score assignments provide a useful framework for anticipating trends in student grades.

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Most popular questions from this chapter

{ For a normal }\end{array}\( distribution, a. Find the \)z\( -score for which a total probability of 0.02 falls more than \)z\( standard deviations (in either direction) from the mean, that is, below \)\mu-z \sigma\( or above \)\mu+z \sigma\( b. For this \)z\(, explain why the probability more than \)z\( standard deviations above the mean equals \)0.01 .\( c. Explain why \)\mu+2.33 \sigma$ is the 99 th percentile.

The juror pool for the upcoming murder trial of a celebrity actor contains the names of 100,000 individuals in the population who may be called for jury duty. The proportion of the available jurors on the population list who are Hispanic is 0.40 . A jury of size 12 is selected at random from the population list of available jurors. Let \(X=\) the number of Hispanics selected to be jurors for this jury. a. Is it reasonable to assume that \(X\) has a binomial distribution? If so, identify the values of \(n\) and \(p\). If not, explain why not. b. Find the probability that no Hispanic is selected. c. If no Hispanic is selected out of a sample of size 12 , does this cast doubt on whether the sampling was truly random? Explain.

In the National Basketball Association, the top free throw shooters usually have probability of about 0.90 of making any given free throw. a. During a game, one such player (Dolph Schayes) shot 10 free throws. Let \(X=\) number of free throws made. What must you assume in order for \(X\) to have a binomial distribution? (Studies have shown that such assumptions are well satisfied for this sport.) b. Specify the values of \(n\) and \(p\) for the binomial distribution of \(X\) in part a. c. Find the probability that he made (i) all 10 free throws and (ii) 9 free throws.

An exit poll is taken of 3000 voters in a statewide election. Let \(X\) denote the number who voted in favor of a special proposition designed to lower property taxes and raise the sales tax. Suppose that in the population, exactly \(50 \%\) voted for it. a. Explain why this scenario would seem to satisfy the three conditions needed to use the binomial distribution. Identify \(n\) and \(p\) for the binomial. b. Find the mean and standard deviation of the probability distribution of \(X\). c. Using the normal distribution approximation, give an interval in which you would expect \(X\) almost certainly to fall, if truly \(p=0.50\). (Hint: You can follow the reasoning of Example 14 on racial profiling.) d. Now, suppose that the exit poll had \(x=1706 .\) What would this suggest to you about the actual value of \(p ?\)

The state of Ohio has several statewide lottery options. One is the Pick 3 game in which you pick one of the 1000 three-digit numbers between 000 and 999\. The lottery selects a three-digit number at random. With a bet of \(\$ 1,\) you win \(\$ 500\) if your number is selected and nothing (\$0) otherwise. (Many states have a very similar type of lottery.) (Source: Background information from www.ohiolottery.com.) a. With a single \(\$ 1\) bet, what is the probability that you win \(\$ 500 ?\) b. Let \(X\) denote your winnings for a \(\$ 1\) bet, so \(x=\$ 0\) or \(x=\$ 500\). Construct the probability distribution for \(X\). c. Show that the mean of the distribution equals 0.50 , corresponding to an expected return of 50 cents for the dollar paid to play. Interpret the mean. d. In Ohio's Pick 4 lottery, you pick one of the 10,000 four-digit numbers between 0000 and 9999 and (with a \(\$ 1\) bet ) win \(\$ 5000\) if you get it correct. In terms of your expected winnings, with which game are you better off - playing Pick 4 , or playing Pick 3 in which you win \(\$ 500\) for a correct choice of a three-digit number? Justify your answer.
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