Question

Suppose a student who knows \(60 \%\) of the material covered in a chapter of a textbook is going to take a five-question objective (each answer is either right or wrong, not multiple choice or true-false) quiz. Let \(X\) be the random variable that gives the number of questions the student answers correctly for each quiz in the sample space of all quizzes the instructor could construct. What is the expected value of the random variable \(X-3 ?\) What is the expected value of \((X-3)^{2}\) ? What is the variance of \(X\) ?

Short answer

The expected value of \(X-3\) is 0, \(E((X-3)^2)\) is 1.2, and the variance of \(X\) is 1.2.

Step-by-step solution

Define the Random Variable X

The student knows 60% of the material, so the probability of answering a question correctly is 0.6, and the probability of answering incorrectly is 0.4. The random variable \(X\) represents the number of questions answered correctly out of 5, so \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = 0.6\).

Calculate the Expected Value of X

For a binomial random variable \(X\), the expected value (mean) is given by \(E(X) = np\). Therefore, the expected value of \(X\) is \(E(X) = 5 \times 0.6 = 3\).

Calculate the Expected Value of X-3

We are interested in \(E(X-3)\). By the properties of expected values, \(E(X-3) = E(X) - 3\). Since we found \(E(X) = 3\), it follows that \(E(X-3) = 3 - 3 = 0\).

Calculate the Expected Value of (X-3)^2

First, note that \((X-3)\) is a transformation of \(X\). The variance can often help in this computation. \(Var(X) = E((X - E(X))^2)\), so \(Var(X - 3) = Var(X)\) due to variance being shift invariant. We need \(E((X-3)^2)\), which can be transformed as \(Var(X) + [E(X-3)]^2\). Since we know \(E(X-3) = 0\), we find \(E((X-3)^2) = Var(X)\).

Calculate the Variance of X

The variance of a binomial random variable is \(Var(X) = np(1-p)\). Therefore, the variance of \(X\) is \(Var(X) = 5 \times 0.6 \times 0.4 = 1.2\).

Use Variance to Find E((X-3)^2)

From Step 4, we have \(E((X-3)^2) = Var(X) + [E(X-3)]^2 = 1.2 + 0 = 1.2\).

Key concepts

Expected Value

In probability, the expected value is a crucial concept that represents the average outcome of a random variable in the long run. Think of it as the mean of a probability distribution, reflecting how we "expect" the variable to behave.
For a binomial distribution, which applies when we have a number of trials with binary outcomes (like right or wrong answers), the expected value is calculated by multiplying the number of trials by the probability of success on one trial.

• For example, in our problem, the number of trials is 5, and the probability of correctly answering a question is 0.6.
• Therefore, the expected value of the number of correct answers is calculated by: \(E(X) = n \cdot p = 5 \times 0.6 = 3\).

Understanding expected value helps in predicting outcomes over time, even though an individual instance may vary from this expectation.

Variance

Variance is a measure of how spread out the numbers in a data set are. In other words, it tells us how much the values differ from the average value (expected value). Highly variable data will have a large variance, while data that is close to the mean will have a small variance.
In the context of the exercise, once we have a binomial random variable, we calculate variance with the formula: \(Var(X) = n \cdot p \cdot (1 - p)\).

• Here, for our student's quiz, the variance is \(Var(X) = 5 \times 0.6 \times 0.4 = 1.2\).

Variance helps in understanding the reliability of the expected value; a smaller variance means the outcomes are more consistently close to the expected value.

Probability

Probability measures the likelihood of a specific outcome occurring within a random event. In simpler terms, it's how often you expect a particular result to happen if you were to repeat the trial over and over.
When dealing with a binomial distribution, the probability structure is set up by the probability of success \(p\) and the number of trials \(n\).

• In our example, the probability of the student answering a question correctly is given as 0.6.
• The opposite event, answering incorrectly, occurs with probability \(1 - p = 0.4\).

This forms the foundation of the student's likelihood of achieving any specific number of correct answers in the quiz.

Random Variable

A random variable represents a numerical outcome of a random process. It's a way to quantify events in probability and statistics. In many scenarios, we are interested in understanding and predicting the behavior of this variable.
In the context of quizzes, the random variable \(X\) represents the number of questions a student answers correctly. Since each answer can either be correct or wrong, and must follow certain probabilities, \(X\) follows what's known as a binomial distribution in this problem.

• Here, the quiz has 5 questions, and the student has a 60% chance to answer each correctly.
• Therefore, \(X\) can take values between 0 and 5, and is characterized by its probability mass function defined for each possible outcome.

Understanding the random variable \(X\) helps us calculate things such as expected outcomes, variance, and the probabilities of different numbers of correct answers.