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Wagga Wagga University has 15.000 students. Let \(X\) be the number of courses for which a randomly chosen student is registered. No student is registered for more than seven courses, and each student is registered for at least one course. The number of students registered for \(i\) courses where \(1 \leq i \leq 7\) is 150,450,1950,3750,5850,2550 , and 300 , respectively. Compute the expected value of the random variable \(X\).

Short Answer

Expert verified
The expected number of courses is 4.57.

Step by step solution

01

Understand the Problem Objective

The objective of this exercise is to calculate the expected value of the random variable \( X \), which represents the number of courses a randomly chosen student at Wagga Wagga University is registered for.
02

List Given Values

The university has a total of 15,000 students. The number of students registered for \( i \) courses is as follows: 150 students for 1 course, 450 for 2 courses, 1950 for 3 courses, 3750 for 4 courses, 5850 for 5 courses, 2550 for 6 courses, and 300 for 7 courses.
03

Calculate Probabilities

Calculate the probability \( P(X = i) \) that a student is registered for \( i \) courses: \[P(X = i) = \frac{\text{Number of students taking } i \text{ courses}}{\text{Total students}}\]For example, \( P(X = 1) = \frac{150}{15000} = 0.01 \). Repeat this for all \( i \).
04

Apply the Formula for Expected Value

The expected value \( E(X) \) of a random variable is calculated using the formula:\[E(X) = \sum_{i=1}^{7} i \cdot P(X = i)\]Substitute the values we calculated in the previous step into this formula to compute the expected value.
05

Perform Calculations

Compute each term in the expected value formula:1. \(1 \cdot 0.01 = 0.01\)2. \(2 \cdot \frac{450}{15000} = 0.06\)3. \(3 \cdot \frac{1950}{15000} = 0.39\)4. \(4 \cdot \frac{3750}{15000} = 1.00\)5. \(5 \cdot \frac{5850}{15000} = 1.95\)6. \(6 \cdot \frac{2550}{15000} = 1.02\)7. \(7 \cdot \frac{300}{15000} = 0.14\)
06

Summarize Result

Add all the calculated terms to find the expected value:\[E(X) = 0.01 + 0.06 + 0.39 + 1.00 + 1.95 + 1.02 + 0.14 = 4.57\]Thus, the expected number of courses that a randomly chosen student is registered for is 4.57.

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Key Concepts

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

Probability Distribution
A probability distribution gives us a mathematical function that shows all the possible values a random variable can take, along with how likely each value is to occur. In simpler terms, it tells us how probabilities are spread across various outcomes.
For instance, in our Wagga Wagga University problem, the random variable \(X\) represents the number of courses a student is registered for. The probability distribution specifies the likelihood of \(X\) being 1, 2, 3, etc., given the total student count is 15,000.
Understanding the probability distribution begins with knowing the frequency of each possible outcome—here, the count of students registered for different numbers of courses—and then translating these frequencies into probabilities. This is done by dividing the number of occurrences (students in a category) by the total number of students (15,000).
  • The key takeaway is that a probability distribution helps us understand the "spread" or "pattern" of a random variable across all its possible values.
Random Variable
A random variable is a mathematical concept used to describe uncertain events numerically.
In the scope of our problem, the random variable \(X\) is defined as the number of courses a randomly selected student is registered for.
This means each student randomly picked from the university's population contributes to a piece of data for the random variable.
  • The values \(X\) can take are between 1 and 7, based on how the students are enrolled in specific number of courses.
  • Random variables can be discrete or continuous. In our example, \(X\) is a discrete random variable because it can take only distinct, separate values (1, 2, 3, etc.).
The concept of a random variable is fundamental in probability and statistics because it allows us to work with and analyze complex random phenomena quantitatively.
Discrete Mathematics
Discrete mathematics is an area of mathematics that deals with countable, distinct entities. It's different from concepts like calculus, which often involve continuous systems.
Our exercise falls under discrete mathematics because it involves a finite set of values—a student can only be enrolled in a whole number of courses from 1 to 7.
  • Discrete mathematics provides the methods needed for calculating probabilities and expected values for discrete random variables, like \(X\).
  • It also encompasses various techniques that allow us to systematically tackle problems involving limited numerical categories.
Learning discrete mathematics offers a robust framework for working with scenarios that involve discrete sets, such as counting, sequencing, and decision-making in finite systems.
Calculation Steps
The computation of the expected value, a form of weighted average, can be approached methodically through carefully structured steps. Here, we outline these steps:

Start by understanding what you need: The goal is to calculate the expected value of the random variable \(X\), representing course enrollment numbers.

Identify your data: Gather the details about how many students are taking different numbers of courses.

  • Translate these numbers into probabilities. For example, if 150 students take 1 course out of 15,000, the probability \(P(X = 1)\) is \(\frac{150}{15000} = 0.01\).
  • Do this for each option up to 7 courses.
After calculating these probabilities, apply the formula for expected value \(E(X) = \sum_{i=1}^{7} i \cdot P(X = i)\).
Calculate each term separately:
  • \(1 \cdot 0.01 = 0.01\)
  • \(2 \cdot 0.03 = 0.06\), and so on.
Sum up all these products to reach the expected value.

Total them all together to get \(E(X) = 4.57\). This result gives an average sense of how many courses a typical student is registered in, showing the mathematical expectation across the student body.

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Most popular questions from this chapter

Suppose that \(\Omega\) is a sample space with a probability density function \(p\), and suppose that \(A \subseteq \Omega\). Let \(P(A)\) denote the probability of \(A\). Assume that \(P(A)>0 .\) Define a function \(p_{1}\) on \(A\) as follows: For \(\omega \in A, p_{1}(\omega)=p(\omega) / P(A)\). (a) Show that if \(\omega_{1}, \omega_{2} \in A\) and \(p\left(\omega_{1}\right), p\left(\omega_{2}\right) \neq 0,\) then \(\frac{p\left(\omega_{1}\right)}{p\left(\omega_{2}\right)}=\frac{p_{1}\left(\omega_{1}\right)}{p_{1}\left(\omega_{2}\right)}\) (b) Show that if \(B\) and \(C\) are nonempty subsets of \(A\) with elements that have positive probabilities, then \(\frac{P(B)}{P(C)}=\frac{P_{1}(B)}{P_{1}(C)}\) (c) Show that \(p_{1}\) is a probability density function on \(\Omega_{1}=A\).

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