Math Problem Statement
Lifetime of a certain hardware is a continuous random variable with density f(x) ={ πΎ β ( π₯ 50) πππ 0 < π₯ < 10 π¦ππππ 0 ππππππππ‘βπππ₯ } (a) Find K. (b) What is the probability of a failure within the first 5 years? (c) What is the expectation of the lifetime?
Solution
Given the problem, we are asked to solve three parts involving a continuous random variable for the lifetime of hardware with the given probability density function .
The probability density function (PDF) is defined as:
\begin{cases} K - \frac{x}{50}, & \text{for } 0 < x < 10 \text{ years} \\ 0, & \text{otherwise} \end{cases}$$ We need to: ### Part (a): Find $$K$$ To find the value of $$K$$, we use the fact that the total probability over the entire support of the PDF must be 1. This means that the integral of $$f(x)$$ over the range from 0 to 10 must equal 1: $$\int_{0}^{10} \left( K - \frac{x}{50} \right) dx = 1$$ Let's calculate this step-by-step: 1. First, integrate each term: $$\int_{0}^{10} K \, dx = K \cdot 10$$ $$\int_{0}^{10} \frac{x}{50} \, dx = \frac{1}{50} \cdot \frac{x^2}{2} \Big|_0^{10} = \frac{1}{50} \cdot \frac{100}{2} = 1$$ 2. So, the total integral becomes: $$K \cdot 10 - 1 = 1$$ 3. Solving for $$K$$: $$10K = 2 \quad \Rightarrow \quad K = \frac{1}{5}$$ Thus, $$K = \frac{1}{5}$$. --- ### Part (b): What is the probability of a failure within the first 5 years? The probability of failure within the first 5 years is the integral of the PDF from 0 to 5: $$P(0 < x < 5) = \int_{0}^{5} \left( \frac{1}{5} - \frac{x}{50} \right) dx$$ 1. Integrate each term: $$\int_{0}^{5} \frac{1}{5} \, dx = \frac{1}{5} \cdot 5 = 1$$ $$\int_{0}^{5} \frac{x}{50} \, dx = \frac{1}{50} \cdot \frac{x^2}{2} \Big|_0^5 = \frac{1}{50} \cdot \frac{25}{2} = 0.25$$ 2. So, the probability is: $$P(0 < x < 5) = 1 - 0.25 = 0.75$$ Thus, the probability of failure within the first 5 years is $$0.75$$ or 75%. --- ### Part (c): What is the expectation of the lifetime? The expectation $$E[X]$$ of a continuous random variable is given by: $$E[X] = \int_{0}^{10} x f(x) \, dx = \int_{0}^{10} x \left( \frac{1}{5} - \frac{x}{50} \right) dx$$ 1. Distribute $$x$$ into the integrand: $$E[X] = \int_{0}^{10} \left( \frac{x}{5} - \frac{x^2}{50} \right) dx$$ 2. Now, integrate each term: $$\int_{0}^{10} \frac{x}{5} \, dx = \frac{1}{5} \cdot \frac{x^2}{2} \Big|_0^{10} = \frac{1}{5} \cdot 50 = 10$$ $$\int_{0}^{10} \frac{x^2}{50} \, dx = \frac{1}{50} \cdot \frac{x^3}{3} \Big|_0^{10} = \frac{1}{50} \cdot \frac{1000}{3} = \frac{20}{3}$$ 3. So, the expected value is: $$E[X] = 10 - \frac{20}{3} = \frac{30}{3} - \frac{20}{3} = \frac{10}{3} \approx 3.33$$ Thus, the expected lifetime is $$\frac{10}{3}$$ years or approximately 3.33 years. --- Would you like more details on any part of this explanation? Here are five related questions you might want to explore: 1. How would the expectation change if the PDF had a different form? 2. What if the lifetime was limited to 15 years instead of 10? 3. Can you calculate the variance of the lifetime distribution? 4. How does the probability of failure differ if the hardware becomes less reliable over time? 5. What is the median lifetime, given this distribution? **Tip:** When finding the expectation of a random variable, make sure to include the variable $$x$$ in the integral, as it's part of the expected value formula.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Probability Density Function (PDF)
Continuous Random Variables
Expectation
Integration
Formulas
Total probability: β« f(x) dx = 1
Probability of failure within a given time: P(a < X < b) = β« f(x) dx from a to b
Expectation of lifetime: E[X] = β« x f(x) dx
Theorems
Fundamental Theorem of Calculus
Expectation of a Random Variable
Suitable Grade Level
Undergraduate Level (Statistics or Probability Course)
Related Recommendation
Probability Density Function and CDF for Electronic Component Lifetime
Finding the Probability Distribution Function from a Given Cumulative Distribution Function
Probability of Component Lifespan Between 14 and 20 Months Using Exponential Distribution
Calculate Spare Components to Ensure 95% System Reliability Over 2000 Hours
Calculating Probability of Death Between Ages 60 and 70 Using PDF