Math Problem Statement

The life (in months) of a certain computer component has a probability density function defined by f(x) = (1/4) * e^(-x/4) for x in [0, ∞). Find the probability that a component randomly selected will last between 14 and 20 months.

Solution

The problem presents a probability density function (PDF) f(x)=14ex/4f(x) = \frac{1}{4} e^{-x/4}, which describes the life (in months) of a certain computer component. We are tasked with finding the probability that the component will last between 14 and 20 months.

To solve this, we need to compute the probability that the random variable XX (representing the life span) falls between 14 and 20. This probability is given by:

P(14X20)=1420f(x)dxP(14 \leq X \leq 20) = \int_{14}^{20} f(x) \, dx

Step 1: Set up the integral

We need to evaluate the following integral:

P(14X20)=142014ex/4dxP(14 \leq X \leq 20) = \int_{14}^{20} \frac{1}{4} e^{-x/4} \, dx

Step 2: Find the antiderivative

The antiderivative of 14ex/4\frac{1}{4} e^{-x/4} is:

14ex/4dx=ex/4+C\int \frac{1}{4} e^{-x/4} \, dx = -e^{-x/4} + C

Step 3: Evaluate the definite integral

Now, we evaluate the antiderivative at the limits 14 and 20:

P(14X20)=[ex/4]1420P(14 \leq X \leq 20) = \left[ -e^{-x/4} \right]_{14}^{20}

This gives:

P(14X20)=e20/4+e14/4P(14 \leq X \leq 20) = -e^{-20/4} + e^{-14/4} P(14X20)=e5+e3.5P(14 \leq X \leq 20) = -e^{-5} + e^{-3.5}

Step 4: Approximate the result

Using approximate values for the exponentials:

e50.0067ande3.50.0302e^{-5} \approx 0.0067 \quad \text{and} \quad e^{-3.5} \approx 0.0302

Therefore:

P(14X20)0.0067+0.0302=0.0235P(14 \leq X \leq 20) \approx -0.0067 + 0.0302 = 0.0235

Thus, the probability that the component lasts between 14 and 20 months is approximately 0.0235, or 2.35%.


Would you like further details or clarifications on any step?

Here are 5 related questions to expand your understanding:

  1. How would you compute the probability for a different time interval, say, between 10 and 30 months?
  2. Can you interpret what the mean life of this component is based on the given probability density function?
  3. What does it mean for a probability density function to be normalized?
  4. How does changing the parameter in the exponential function affect the component’s life expectancy?
  5. How could you use this model to find the median life expectancy of the component?

Tip: When working with exponential distributions, the rate parameter determines both the scale and the average life expectancy of the system being modeled.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Exponential Distribution
Integral Calculus

Formulas

f(x) = (1/4) * e^(-x/4)
P(a ≤ X ≤ b) = ∫[a,b] f(x) dx

Theorems

Exponential Distribution
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Calculus and Probability)