Math Problem Statement

What proportion of students take between 1.5 and 2.0 hours to finish the exam? What is the mean time for students to complete the exam?

Solution

Let's start by solving part (a) and then move on to part (b) based on the provided probability density function (PDF) and the questions.

Given PDF:

\frac{x^3}{4} & \text{if } 0 < x < 2, \\ 0 & \text{otherwise}. \end{cases}$$ #### Part (a) - Proportion of students taking between 1.5 and 2.0 hours: This proportion corresponds to the probability that $$x$$ is between 1.5 and 2.0. We need to integrate the given PDF over this interval: $$P(1.5 \leq x \leq 2) = \int_{1.5}^{2} \frac{x^3}{4} \, dx$$ I'll calculate this integral.The proportion of students who take between 1.5 and 2.0 hours to finish the exam is approximately **0.684** (rounded to three decimal places). #### Part (b) - Mean time for students to complete the exam: The mean (or expected value) for the time $$x$$ is given by the formula: $$E(x) = \int_{0}^{2} x \cdot p(x) \, dx = \int_{0}^{2} x \cdot \frac{x^3}{4} \, dx = \int_{0}^{2} \frac{x^4}{4} \, dx$$ Let's calculate this integral to find the mean time.The mean time for students to complete the exam is **1.6 hours** (rounded to one decimal place). Would you like further explanation on any of these calculations, or do you have any other questions? Here are 5 related questions to consider: 1. How would you calculate the variance of the time it takes for students to complete the exam? 2. What is the cumulative distribution function (CDF) for this PDF? 3. How can we calculate the median time for students to complete the exam? 4. What proportion of students take less than 1 hour to complete the exam? 5. How would the PDF change if the maximum completion time were 3 hours instead of 2? **Tip**: The expected value (mean) of a continuous random variable provides a long-run average outcome of the variable over many observations.

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Math Problem Analysis

Mathematical Concepts

Probability
Integration
Expected Value
Continuous Probability Distribution

Formulas

p(x) = x^3 / 4 for 0 < x < 2
Proportion: P(1.5 ≤ x ≤ 2) = ∫(1.5 to 2) (x^3 / 4) dx
Mean: E(x) = ∫(0 to 2) x * (x^3 / 4) dx

Theorems

Fundamental Theorem of Calculus
Expected Value Formula for Continuous Distributions

Suitable Grade Level

Undergraduate (Probability/Statistics)