E. Chapter 5 HW - Continuous Random Variables Curre *A, Category: Homework 5/50 1 $$ (2) $$ 3 4. 5 6 7 8 9 10 Current learning objective: Continuous Probability Functions Question 2 Practice similar questions

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Sure, I'd be happy to help you with Chapter 5 on Continuous Random Variables, specifically focusing on Continuous Probability Functions. Since the exact details of **Question 2** aren't provided, I'll outline a typical Problem 2 scenario in this topic, walk you through its solution, and then provide similar practice questions for further practice. --- ### **Sample Question 2** **Problem:** Suppose the lifetime $$ X $$ of a certain battery (in hours) is a continuous random variable with the probability density function (pdf): $$ f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{for } x \geq 0, \ 0 & \text{otherwise}. \end{cases} $$ Here, $$ \lambda = 0.5 $$ per hour. **Tasks:** a) **Find the probability that a battery lasts at least 4 hours.** b) **Determine the expected lifetime (mean) of the battery.** c) **Find the probability that a battery lasts between 2 and 5 hours.** --- **Solution:** This pdf represents an **Exponential Distribution**, commonly used to model time until an event occurs (e.g., battery failure). **a) Probability that a battery lasts at least 4 hours ($$ P(X \geq 4) $$):** $$ P(X \geq 4) = 1 - P(X < 4) = e^{-\lambda \cdot 4} = e^{-0.5 \cdot 4} = e^{-2} \approx 0.1353 $$ **b) Expected lifetime (mean) of the battery:** For an Exponential Distribution, the mean $$ \mu $$ is: $$ \mu = \frac{1}{\lambda} = \frac{1}{0.5} = 2 \text{ hours} $$ **c) Probability that a battery lasts between 2 and 5 hours ($$ P(2 \leq X \leq 5) $$):** $$ P(2 \leq X \leq 5) = P(X \leq 5) - P(X < 2) = \left(1 - e^{-\lambda \cdot 5}\right) - \left(1 - e^{-\lambda \cdot 2}\right) = e^{-0.5 \cdot 2} - e^{-0.5 \cdot 5} = e^{-1} - e^{-2.5} \approx 0.3679 - 0.0821 = 0.2858 $$ --- ### **Similar Practice Questions** Here are some practice problems similar to the one above to help reinforce your understanding of Continuous Probability Functions: --- **Practice Question 1:** Let $$ Y $$ be a continuous random variable representing the time (in minutes) it takes for a customer to be served at a bank. The probability density function of $$ Y $$ is given by: $$ f(y) = \begin{cases} 3e^{-3y} & \text{for } y \geq 0, \ 0 & \text{otherwise}. \end{cases} $$ **Tasks:** a) Calculate $$ P(Y \leq 2) $$, the probability that a customer is served within 2 minutes. b) Find the mean service time. c) Determine $$ P(1 \leq Y \leq 4) $$, the probability that a customer's service time is between 1 and 4 minutes. --- **Practice Question 2:** A continuous random variable $$ Z $$ has the probability density function: $$ f(z) = \begin{cases} \frac{1}{5} & \text{for } 0 \leq z \leq 5, \ 0 & \text{otherwise}. \end{cases} $$ **Tasks:** a) Find $$ P(Z > 3) $$. b) Determine the expected value $$ E[Z] $$. c) Compute $$ P(2 \leq Z \leq 4) $$. --- **Practice Question 3:** The height $$ H $$ (in centimeters) of adult males in a certain population is modeled by the following probability density function: $$ f(h) = \begin{cases} \frac{h}{100^2} & \text{for } 0 \leq h \leq 100, \ 0 & \text{otherwise}. \end{cases} $$ **Tasks:** a) Verify that $$ f(h) $$ is a valid probability density function. b) Find $$ P(H \geq 75) $$. c) Calculate the mean height $$ E[H] $$. --- **Practice Question 4:** A machine produces widgets, and the time $$ T $$ (in hours) between successive widgets being produced follows an exponential distribution with parameter $$ \lambda = 0.3 $$. **Tasks:** a) What is the probability that the machine takes more than 5 hours to produce the next widget? b) What is the expected time between widget productions? c) Find the probability that the machine produces a widget between 2 and 4 hours from now. --- ### **Tips for Solving Continuous Probability Function Problems** 1. **Understand the Distribution:** - Identify whether the problem deals with common distributions like Exponential, Uniform, Normal, etc. 2. **Verify the PDF:** - Ensure that the total area under the pdf curve equals 1. - For non-standard distributions, integrate the pdf over its domain to check. 3. **Use Relevant Formulas:** - **Exponential Distribution:** - $$ P(X \leq x) = 1 - e^{-\lambda x} $$ - Mean $$ \mu = \frac{1}{\lambda} $$ - **Uniform Distribution:** - $$ f(x) = \frac{1}{b - a} $$ for $$ a \leq x \leq b $$ - Mean $$ \mu = \frac{a + b}{2} $$ 4. **Calculate Probabilities:** - Probabilities are found by integrating the pdf over the desired interval. - For cumulative probabilities, use the cumulative distribution function (CDF). 5. **Practice Integration:** - Be comfortable with basic integration techniques as they are essential for finding areas under curves. 6. **Check Units:** - Ensure that all units are consistent, especially when dealing with rates (like $$ \lambda $$) and time. --- Feel free to attempt these practice questions, and let me know if you need solutions or further explanations for any of them! **Sample Question 2:** *Problem:* Suppose the lifetime $$ X $$ of a battery (in hours) is a continuous random variable with the probability density function (pdf): $$ f(x) = \begin{cases} 0.5 e^{-0.5 x} & \text{for } x \geq 0, \ 0 & \text{otherwise}. \end{cases} $$ **Tasks:** a) Find the probability that a battery lasts at least 4 hours. b) Determine the expected lifetime of the battery. c) Find the probability that a battery lasts between 2 and 5 hours. --- **Solution:** a) $$ P(X \geq 4) = e^{-0.5 \times 4} = e^{-2} \approx 0.1353 $$ b) $$ \mu = \frac{1}{0.5} = 2 $$ hours c) $$ P(2 \leq X \leq 5) = e^{-0.5 \times 2} - e^{-0.5 \times 5} = e^{-1} - e^{-2.5} \approx 0.3679 - 0.0821 = 0.2858 $$ --- **Practice Questions:** 1. **Practice Question 1:** - **a)** $$ P(Y \leq 2) = e^{-3 \times 2} = e^{-6} \approx 0.0025 $$ - **b)** $$ \mu = \frac{1}{3} $$ minutes - **c)** $$ P(1 \leq Y \leq 4) = e^{-3 \times 1} - e^{-3 \times 4} = e^{-3} - e^{-12} \approx 0.0498 - 0.00000614 = 0.0498 $$ 2. **Practice Question 2:** - **a)** $$ P(Z > 3) = 1 - \frac{3}{5} = 0.4 $$ - **b)** $$ E[Z] = \frac{5}{2} = 2.5 $$ minutes - **c)** $$ P(2 \leq Z \leq 4) = \frac{4 - 2}{5} = 0.4 $$ 3. **Practice Question 3:** - **a)** Verify that $$ \int_{0}^{100} \frac{h}{100^2} \, dh = \frac{1}{2} $$, which is less than 1. To make it a valid pdf, multiply by 2. - **b)** $$ P(H \geq 75) = \int_{75}^{100} \frac{2h}{100^2} \, dh = \frac{2}{100^2} \times \frac{100^2 - 75^2}{2} = \frac{100^2 - 75^2}{100^2} = \frac{5625}{10000} = 0.5625 $$ - **c)** $$ E[H] = \frac{2 \times (0 + 100)}{2} = 100 $$ cm 4. **Practice Question 4:** - **a)** $$ P(T > 5) = e^{-0.3 \times 5} = e^{-1.5} \approx 0.2231 $$ - **b)** $$ \mu = \frac{1}{0.3} \approx 3.333 $$ hours - **c)** $$ P(2 \leq T \leq 4) = e^{-0.3 \times 2} - e^{-0.3 \times 4} = e^{-0.6} - e^{-1.2} \approx 0.5488 - 0.3012 = 0.2476 $$ **Tips:** - Identify the distribution type. - Verify the pdf is valid. - Use integration for probabilities. - Ensure units are consistent. - Practice with different distributions.

E. Chapter 5 HW - Continuous Random Variables Curre *A, Category: Homework 5/50 1 \( (2) \) 3 4. 5 6 7 8 9 10 Current learning objective: Continuous Probability Functions Question 2 Practice similar questions

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