Math Problem Statement

The length of time in minutes, X, between 911 calls in Metropolis has an exponential distribution with a mean of 20.00 minutes. Thus, the density of X is:

f(x,𝜆) = 𝜆 e−𝜆x

 for 0 ≤ x ≤ ∞, 𝜆 is what R calls rate.

Hint: This is a problem involving the exponential distribution. Knowing the parameter 𝜆 for the distribution allows you to easily answer parts a ,b ,c and use the built-in R functions for the exponential distribution (dexp(), pexp(), qexp()) for other parts . Or (not recommended) you should be able to use the R integrate command with f(x) defined as above or with dexp() for all parts.

a) What is the value of 𝜆? 

b) What is the variance of X? 

c) What is the standard deviation of X? 

d) What is the probability that X is greater than its expected value? 

e) What is the probability that X is > 25? 

f) What is the probability that X is > 50? 

g) What is the probability that X > 50 given that X > 25? 

h) What is the median of X? 

l) Enter any comments in the text box below.

Solution

To solve this problem involving the exponential distribution, let's go through each part step by step.

Given Information

The mean of the exponential distribution μ\mu is 20.00 minutes. The rate parameter λ\lambda is related to the mean by the formula:

λ=1μ\lambda = \frac{1}{\mu}

a) What is the value of λ\lambda?

Using the mean:

λ=120=0.05\lambda = \frac{1}{20} = 0.05

b) What is the variance of XX?

The variance σ2\sigma^2 of an exponential distribution is given by:

σ2=1λ2\sigma^2 = \frac{1}{\lambda^2}

Substituting the value of λ\lambda:

σ2=1(0.05)2=10.0025=400\sigma^2 = \frac{1}{(0.05)^2} = \frac{1}{0.0025} = 400

c) What is the standard deviation of XX?

The standard deviation σ\sigma is the square root of the variance:

σ=400=20\sigma = \sqrt{400} = 20

d) What is the probability that XX is greater than its expected value?

The probability P(X>μ)P(X > \mu) for an exponential distribution is calculated as follows:

P(X>μ)=P(X>20)=eλ20=e0.0520=e10.3679P(X > \mu) = P(X > 20) = e^{-\lambda \cdot 20} = e^{-0.05 \cdot 20} = e^{-1} \approx 0.3679

e) What is the probability that X>25X > 25?

Using the exponential distribution's survival function:

P(X>25)=eλ25=e0.0525=e1.250.2865P(X > 25) = e^{-\lambda \cdot 25} = e^{-0.05 \cdot 25} = e^{-1.25} \approx 0.2865

f) What is the probability that X>50X > 50?

P(X>50)=eλ50=e0.0550=e2.50.0821P(X > 50) = e^{-\lambda \cdot 50} = e^{-0.05 \cdot 50} = e^{-2.5} \approx 0.0821

g) What is the probability that X>50X > 50 given that X>25X > 25?

Using the memoryless property of the exponential distribution:

P(X>50X>25)=P(X>5025)=P(X>25)=e0.05250.2865P(X > 50 | X > 25) = P(X > 50 - 25) = P(X > 25) = e^{-0.05 \cdot 25} \approx 0.2865

h) What is the median of XX?

The median MM of an exponential distribution can be calculated as:

M=ln(2)λM = \frac{\ln(2)}{\lambda}

Calculating:

M0.69310.0513.86M \approx \frac{0.6931}{0.05} \approx 13.86

Summary of Answers

  • a) λ=0.05\lambda = 0.05
  • b) Variance σ2=400\sigma^2 = 400
  • c) Standard Deviation σ=20\sigma = 20
  • d) P(X>20)0.3679P(X > 20) \approx 0.3679
  • e) P(X>25)0.2865P(X > 25) \approx 0.2865
  • f) P(X>50)0.0821P(X > 50) \approx 0.0821
  • g) P(X>50X>25)0.2865P(X > 50 | X > 25) \approx 0.2865
  • h) Median 13.86\approx 13.86

If you have any questions or need further clarification on any part, let me know! Here are some follow-up questions:

  1. How is the exponential distribution applied in real-world scenarios?
  2. Can you explain the memoryless property in more detail?
  3. What would happen to the probabilities if the mean was decreased?
  4. How can we graph the probability density function for this distribution?
  5. Are there any other distributions that have similar properties to the exponential distribution?

Tip: Remember that the exponential distribution is often used to model time until an event occurs, such as failures or arrivals.

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Math Problem Analysis

Mathematical Concepts

Exponential Distribution
Probability
Statistics

Formulas

Rate (𝜆) = 1 / Mean (𝜇)
Variance = 1 / 𝜆^2
Standard Deviation = sqrt(Variance)
P(X > x) = e^(-𝜆 x)
Median = ln(2) / 𝜆

Theorems

Memoryless Property of Exponential Distribution

Suitable Grade Level

College Statistics or Advanced High School