Problem 10 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 7 through \(10, f(x)\) is a probability density function for a particular continuous random variable \(X\). In each case, find the indicated probabilities. $$ \begin{aligned} &f(x)= \begin{cases}0.25 x e^{-x / 2} & \text { if } x \geq 0 \\ 0 & \text { otherwise }\end{cases} \\ &P(X \geq 2) \text { and } P(0 \leq X \leq 3) \end{aligned} $$

Step-by-Step Solution

Answer

P(X ≥ 2) = 0.125\(e^{-1}\) and P(0 ≤ X ≤ 3) = 0.125\(e^{-3 / 2} - 1\).

Step 1: Identify the given function

The given probability density function (pdf) is \( f(x) = 0.25 x e^{-x / 2} \) for \( x \geq 0 \), and zero otherwise \( f(x) = 0 \) for \( x < 0 \).

Step 2: Expression for P(X ≥ 2)

To find \( P(X \geq 2) \), calculate the integral of \( f(x) \) from 2 to infinity:\[ P(X \geq 2) = \int_{2}^{\infty} 0.25 x e^{-x / 2} \, dx \]

Step 3: Compute the integral for P(X ≥ 2)

Use integration by parts where \( u = x \) and \( dv = 0.25 e^{-x / 2} dx \). Thus, \( du = dx \) and \( v = -0.5 e^{-x / 2} \):\[ \int 0.25 x e^{-x/2} \, dx = 0.25 \left( -0.5 x e^{-x/2} - \int -0.5 e^{-x/2} dx \right) = -0.125 x e^{-x / 2} + 0.125 e^{-x / 2} \]Evaluating from 2 to \( \infty \):\[ \left[ -0.125 x e^{-x / 2} + 0.125 e^{-x / 2} \right]_{2}^{\infty} = 0 - \left( -0.125(2) e^{-1} + 0.125 e^{-1} \right) = 0.125 e^{-1} \]

Step 4: Expression for P(0 ≤ X ≤ 3)

To find \( P(0 \leq X \leq 3) \), calculate the integral of \( f(x) \) from 0 to 3:\[ P(0 \leq X \leq 3) = \int_{0}^{3} 0.25 x e^{-x / 2} \, dx \]

Step 5: Compute the integral for P(0 ≤ X ≤ 3)

Using integration by parts similarly as before:\[ \int 0.25 x e^{-x/2} \, dx = -0.125 x e^{-x / 2} + 0.125 e^{-x/2} \]Evaluating from 0 to 3:\[ \left[ -0.125 x e^{-x / 2} + 0.125 e^{-x / 2} \right]_{0}^{3} = \left( -0.125(3) e^{-3 / 2} + 0.125 e^{-3 / 2} \right) - \left( 0 + 0.125 \right) = 0.125 \left( e^{-3 / 2} - 1 \right) \]

Key Concepts

Continuous Random Variable

Understanding continuous random variables is key to solving probability problems with probability density functions (pdf).
A continuous random variable can take any value within a given range. Unlike discrete variables, it is not limited to distinct, separate values.
In these exercises, the variable X is continuous, as it represents a range of outcomes that can be measured precisely.

The pdf associated with a continuous random variable gives the density of the probability distribution over a range of values. For example, for 0 ≤ x, the function is given by:
\( f(x) = 0.25x \ e^{-x/2} \).
The total area under this curve equals 1, ensuring that the sum of all probabilities in the range is 1.

Integration by Parts

The solution to these probability problems often involves the technique known as integration by parts.
It's a vital tool in calculus for solving integrals of products of functions.
The formula is: \[ \int u \, dv = uv - \int v \, du \.\]
It helps when dealing with the integral of a function that is a product of two simpler functions.

In our solution, we need to find integrals of expressions like \(0.25x e^{-x/2} \). Breaking it down:

• Choose \( u = x \) and \( dv = 0.25 e^{-x/2} dx \).
• Then, differentiate \( u \) to get \( du = dx \).
• Integrate \( dv \) to find \( v = -0.5 e^{-x/2} \).

Combining these through the integration by parts formula allows us to solve the integrals required for the probability calculations.

Probability Calculation

Calculating probabilities for continuous random variables involves integrals of their pdfs.
For example, to find the probability that X is greater than or equal to 2, we integrate the pdf from 2 to infinity.
\[ P(X \geq 2) = \int_{2}^{\infty} 0.25 x e^{-x / 2} \, dx \.\]
Similarly, to find the probability that X falls between 0 and 3, we integrate the pdf from 0 to 3:
\[ P(0 \leq X \leq 3) = \int_{0}^{3} 0.25 x e^{-x / 2} \, dx \.\]

Integrals of the pdf over specified ranges give the probability that the random variable X will fall within those ranges.
These computations are crucial for understanding how likely different outcomes are, based on the given probability distribution.

Applied Calculus

Applied Calculus plays a significant role in understanding and analyzing continuous random variables.
It involves using integration techniques to handle real-world problems where outcomes are not discrete but continuous.

Here's a quick overview of the steps involved in these probability calculations:

• Identify the function provided, ensuring it meets the conditions of a PDF (non-negative and integrates to 1 over the range).
• Set up the appropriate integral for the probability you need to calculate.
• Apply integration techniques, specifically, integration by parts for complex functions.

This analytical approach provided by Applied Calculus equips students to solve problems involving probabilities and random events with precision.
Understanding these techniques and their applications ensures a solid foundation for further studies in statistics and mathematics.