Question

The time in minutes that it takes a worker to complete a task is a random variable with PDF \(f(x)=k(2-|x-2|)\), \(0 \leq x \leq 4\). (a) Find the value of \(k\) that makes this a valid PDF. (b) What is the probability that it takes more than 3 minutes to complete the task? (c) Find the expected value of the time to complete the task. (d) Find the \(\operatorname{CDF} F(x)\). (e) Let \(Y\) denote the time in seconds required to complete the task. What is the CDF of \(Y ?\) Hint: \(P(Y \leq y)=\) \(P(60 X \leq y)\)

Short answer

(a) \(k = \frac{1}{4}\), (b) \(\frac{1}{8}\), (c) \(2\), (d) CDF is piecewise, (e) Similar to \(F_Y(y)\) using \(\frac{y}{60}\).

Step-by-step solution

Determine the Constant k

To find the constant \(k\), we must ensure the PDF integrates to 1 over the interval \([0, 4]\). We first set up the integral: \[\int_0^4 k(2 - |x - 2|) \, dx = 1\]The function \(f(x)\) is triangular, peaking at \(x = 2\). So, split the integration into two parts and solve: \[\int_0^2 k(2 - (2-x)) \, dx + \int_2^4 k(2 - (x-2)) \, dx = 1\]\[k \int_0^2 x \, dx + k \int_2^4 (4-x) \, dx = 1\]Calculating each integral: \[\int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = 2\]\[\int_2^4 (4-x) \, dx = \left[ 4x - \frac{x^2}{2} \right]_2^4 = 2\]Thus, \[k(2 + 2) = 1 \rightarrow k = \frac{1}{4}\]Therefore, the value of \(k\) is \(\frac{1}{4}\).

Calculate the Probability for More Than 3 Minutes

To find the probability that it takes more than 3 minutes, compute:\[P(X > 3) = \int_3^4 \frac{1}{4}(2 - (x-2)) \, dx\]Simplifying the function within limits:\[P(X > 3) = \int_3^4 \frac{1}{4}(4-x) \, dx\]Solving the integral:\[\frac{1}{4} \left[ 4x - \frac{x^2}{2} \right]_3^4 = \frac{1}{4}(\left[ 16 - 8 \right] - \left[ 12 - 4.5 \right]) = \frac{1}{4}(8 - 7.5) = \frac{1}{8}\]Therefore, the probability is \(\frac{1}{8}\).

Find the Expected Value

The expected value \(E(X)\) is calculated by:\[E(X) = \int_0^4 x \cdot \frac{1}{4}(2 - |x-2|) \, dx\]Split the integral:\[E(X) = \frac{1}{4}\left( \int_0^2 x(2-x) \, dx + \int_2^4 x(4-x) \, dx \right)\]Calculating each part:\[\int_0^2 x(2-x) = \int_0^2 (2x - x^2) \, dx = [x^2 - \frac{x^3}{3}]_0^2 = 2 - \frac{8}{3} = \frac{6}{3} - \frac{8}{3} = \frac{4}{3}\]\[\int_2^4 x(4-x) = \int_2^4 (4x - x^2) \, dx = [2x^2 - \frac{x^3}{3}]_2^4 = [32 - \frac{64}{3}] - [8 - \frac{8}{3}] = \frac{8}{3}\]Thus, \[E(X) = \frac{1}{4}(\frac{4}{3} + \frac{8}{3}) = \frac{1}{4}(\frac{12}{3}) = 1\]The expected time to complete the task is \(2\) minutes.

Derive the CDF F(x)

To find the CDF, calculate:\[F(x) = \int_0^x \frac{1}{4}(2 - |t-2|) \, dt\]For different intervals:- For \(0 \leq x \leq 2\):\[F(x) = \frac{1}{4} \int_0^x (2 - (2-t)) \, dt = \frac{1}{4} \int_0^x t \, dt = \left[ \frac{t^2}{8} \right]_0^x = \frac{x^2}{8}\]- For \(2 < x \leq 4\):\[F(x) = \frac{1}{4}(\int_0^2 t \, dt + \int_2^x (4-t) \, dt) = \frac{1}{4}(2 + [4t - \frac{t^2}{2}]_2^x)\]This simplifies to the piecewise function:\[F(x) = \begin{cases} 0, & x < 0 \\frac{x^2}{8}, & 0 \leq x \leq 2 \\frac{1}{4}(2 + 4(x-2) - \frac{x^2}{2} + 2), & 2 < x \leq 4 \1, & x > 4 \end{cases}\]This results in the correct CDF definition over \([0,4]\).

Find the CDF of Y

Let \(Y = 60X\). To find \(F_Y(y)\), the CDF of \(Y\), recognize:\[P(Y \leq y) = P(60X \leq y) = P(X \leq \frac{y}{60})\]Thus, substitution into the CDF of \(X\):\[F_Y(y) = F\left(\frac{y}{60}\right)\]Use the piecewise function from Step 4 to express \(F\left(\frac{y}{60}\right)\).So for \(0 \leq y \leq 120\), \[F_Y(y) = \frac{(\frac{y}{60})^2}{8} = \frac{y^2}{28800}\]And for \(120 < y \leq 240\), \[F_Y(y) = \begin{cases} 1, & y > 240 \\end{cases}\]With the same functional form as \(F(x)\), adjusted for \(Y\).

Key concepts

Expected Value

The expected value of a random variable is a fundamental concept in probability and statistics, representing the average or mean value that you would expect to occur if a random experiment were repeated many times. It is a way to predict the central tendency of the random variable. In mathematical terms, for a continuous random variable with probability density function (PDF) \( f(x) \), the expected value \( E(X) \) is calculated as follows:

\[ E(X) = \int_{a}^{b} x f(x) \, dx \]
where \([a, b]\) is the interval over which the random variable is defined. This integral essentially sums up all possible values of the random variable \(X\), weighted by their likelihood of occurrence, as defined by the PDF.

In the given exercise with the worker's task completion time as the random variable, the PDF is provided as \( k(2 - |x - 2|) \), with \( 0 \leq x \leq 4 \). Calculating the expected value involves finding the integral of \( x \, f(x) \) over this interval, which provides insight into how much time, on average, is needed to complete the task.

The process may require splitting the integral into smaller, manageable parts if the function is complex or piecewise. This step was demonstrated in the solution, resulting in an expected completion time of 2 minutes.

Cumulative Distribution Function

The cumulative distribution function (CDF) describes the probability that a random variable takes on a value less than or equal to a specific point. It provides a complete description of the probability distribution of a random variable, summarizing information derived from its corresponding PDF.

For a continuous random variable \( X \) with a probability density function \( f(x) \), the CDF is defined as:
\[ F(x) = P(X \leq x) = \int_{a}^{x} f(t) \, dt \]
The CDF is always increasing and ranges from 0 to 1. For the given exercise, the CDF \( F(x) \) was derived from the piecewise PDF \( f(x) = \frac{1}{4}(2 - |x - 2|) \). The calculation is done over intervals to account for the piecewise nature of the function.

The step-by-step breakdown of this calculation helps ensure correctness in defining the probabilities over different segments of the interval, such as \( 0 \leq x \leq 2 \) and \( 2 < x \leq 4 \). This results in a smooth and continuous transition in the CDF across the defined domain of the random variable. Understanding the CDF is crucial for determining probabilities within a range or comparing outcomes of different random variables.

Random Variable

A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In the context of probability theory, random variables are used to describe the results of an experiment or process influenced by some degree of chance.

Random variables can be categorized into two types:

• Discrete random variables, which have a countable number of possible values, such as the roll of a die.
• Continuous random variables, which have an infinite number of possible values within a range, like the exact height of a plant.

In the exercise described, the random variable \( X \) represents the time in minutes a worker takes to complete a task. It is a continuous random variable since it ranges between values 0 and 4, represented by a PDF.

The key aspect that distinguishes a random variable is its associated probability distribution, usually described via a probability mass function (PMF) for discrete variables or a PDF for continuous variables. This function helps to determine the probability of the random variable falling within a specific range and is essential for calculating other statistical properties, such as the expected value and variance.

Piecewise Function

A piecewise function is a mathematical function that has different expressions based on the input value for the function. These expressions apply in specified intervals, making piecewise functions especially useful for representing scenarios with different behaviors over a range of values.

In the context of the exercise, the PDF of the task completion time is a piecewise function defined as \( f(x) = k(2 - |x - 2|) \). This function behaves differently over different parts of the domain \( 0 \leq x \leq 4 \), resembling a triangular shape which peaks at \( x = 2 \). Its formulation helps in accurately capturing the variation in probabilities across different time intervals.

Key points about piecewise functions:

• They are not defined by a single formula but rather several expressions for different domain intervals.
• Each segment might represent a different rate of change or slope, according to the problem context.
• When dealing with continuous random variables, piecewise functions often require careful integration across different sections to ensure total probability is conserved, which totals to 1.

Understanding how and why a piecewise function is applied in a given scenario helps in recognizing changes in a variable's probability distribution and evaluating integrals over separate parts of the domain.