Problem 7 - Applied Calculus

Question

Suppose that \(x\) measures the time (in hours) it takes for a student to complete an exam. All students are done within two hours and the density function for \(x\) is $$ p(x)=\left\\{\begin{array}{ll} x^{3} / 4 & \text { if } 0

Step-by-Step Solution

Answer

(a) 28.1%, (b) Mean = 1.6 hours, (c) Median ≈ 1.68 hours.

Step 1: Understanding the Problem

The problem involves a piecewise probability density function (PDF) \( p(x) \), defined over the interval \( 0 < x < 2 \) with \( p(x) = \frac{x^3}{4} \). We need to solve for the proportion, mean, and median of the distribution given this function.

Step 1: Proportion Calculation

To find the proportion of students taking between 1.5 and 2.0 hours, we need to integrate the PDF \( p(x) = \frac{x^3}{4} \) from \( x = 1.5 \) to \( x = 2.0 \). \[\text{Proportion} = \int_{1.5}^{2.0} \frac{x^3}{4} \, dx\] Calculating, we get: \[ \int \frac{x^3}{4} \, dx = \frac{x^4}{16} \]Thus, we evaluate:\[\left. \frac{x^4}{16} \right|_{1.5}^{2.0} = \frac{16}{16} - \frac{5.0625}{16} = \frac{10.9375}{16} = 0.6836\]

Step 2: Mean Calculation

The mean time for students to complete the exam is calculated using the formula for the expected value of a continuous random variable. \[ \text{Mean (}\mu\text{)} = \int_{0}^{2} x \cdot p(x) \, dx = \int_{0}^{2} x \cdot \frac{x^3}{4} \, dx = \int_{0}^{2} \frac{x^4}{4} \, dx \]\[= \left. \frac{x^5}{20} \right|_{0}^{2} = \frac{32}{20} = 1.6\]

Step 3: Median Calculation

To find the median, \( m \), we solve for \( F(m) = 0.5 \), where \( F(x) \) is the cumulative distribution function (CDF). The CDF of \( x \) is:\[ F(x) = \int_{0}^{x} \frac{t^3}{4} \, dt = \left. \frac{t^4}{16} \right|_{0}^{x} = \frac{x^4}{16} \]Setting the CDF equal to 0.5:\[ \frac{m^4}{16} = 0.5 \rightarrow m^4 = 8 \rightarrow m = 8^{1/4} \rightarrow m \approx 1.68 \]

Step 5: Check and Review

Verify that each step is properly calculated and that intermediate steps are consistent with the expectations of the problem.

Key Concepts

Mean Calculation

The mean, often referred to as the expected value, gives us an average measure of the time it takes for a student to complete the exam. It represents where the "center" of the data's distribution lies and helps us understand the typical completion time for the majority of students.

To find the mean, we use the formula for the expected value of a continuous random variable, expressed as follows:

• \( ext{Mean (}\,\mu\,) = \int_{a}^{b} x \cdot p(x) \, dx \)

For this particular scenario, the PDF is defined as \( p(x) = \frac{x^3}{4} \) over the interval \( 0 < x < 2 \). Therefore, we calculate:

• \[ \text{Mean } (\mu) = \int_{0}^{2} x \cdot \frac{x^3}{4} \, dx = \int_{0}^{2} \frac{x^4}{4} \, dx \]
• \[ = \left. \frac{x^5}{20} \right|_{0}^{2} \]
• \[ = \frac{32}{20} = 1.6 \]

Thus, the mean completion time is 1.6 hours. This implies that on average, students spend 1.6 hours to finish their exam.

Median Calculation

The median provides a central value of the data set such that half of the students complete the exam below this time and half finish above it. In probability, the median is found using the cumulative distribution function (CDF).

To determine the median, we solve for the time \( m \) where the CDF equals 0.5. The CDF is obtained through:

• \[ F(x) = \int_{0}^{x} \frac{t^3}{4} \, dt \]
• \[ = \left. \frac{t^4}{16} \right|_{0}^{x} = \frac{x^4}{16} \]

By setting the CDF equal to 0.5, we find:

• \[ \frac{m^4}{16} = 0.5 \]

Thus, \( m^4 = 8 \), leading to \( m = 8^{1/4} \) or approximately 1.68. Therefore, half of the students will finish the exam in less than 1.68 hours and the other half will take longer.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a random variable represents the probability that the variable takes a value less than or equal to a certain number. It offers a comprehensive view of the likelihood for different outcomes and is essential for calculating thresholds like the median.

For our given probability density function \( p(x) = \frac{x^3}{4} \), the CDF is computed as:

• \[ F(x) = \int_{0}^{x} \frac{t^3}{4} \, dt \]
• \[ = \left. \frac{t^4}{16} \right|_{0}^{x} = \frac{x^4}{16} \]

This CDF function helps us understand how the probability accumulates over the interval from zero to any point \( x \) within the range. It's crucial for any calculations involving proportions or to find specific probability-related values like percentiles or medians.