Probability - Poisson Random Variable?

Click For Summary
SUMMARY

The random variable X representing the number of potholes in a 30-mile stretch of I80 during a Pennsylvania winter follows a Poisson distribution. The expected value and variance of X are both calculated as λ = 1.6 potholes per 10 miles, leading to an expected value of 4.8 potholes for 30 miles. The pothole repair expense Y, calculated as Y = 5000 * X, has an expected value of $24,000 and a variance of $5000² * λ, which simplifies to $25,000,000.

PREREQUISITES
  • Understanding of Poisson distribution and its properties
  • Knowledge of expected value and variance calculations
  • Familiarity with basic probability concepts
  • Ability to apply linear transformations to random variables
NEXT STEPS
  • Study the properties of Poisson distribution in detail
  • Learn how to compute expected value and variance for transformed random variables
  • Explore real-world applications of Poisson processes in maintenance and repair scenarios
  • Investigate the implications of variance in financial forecasting for repair costs
USEFUL FOR

Students in statistics, data analysts, and professionals involved in maintenance planning and budgeting for infrastructure repairs.

tjackson
Messages
4
Reaction score
0
1. Homework Statement

During a typical Pennsylvania winter, I80 averages 1.6 potholes per 10 miles. A certain county is responsible for repairing potholes in a 30 mile stretch of the interstate. Let X denote the number of potholes the county will have to repair at the end of next winter.
1. The random variable X is

(i) binomial (ii) hypergeometric (iii) negative binomial (iv) Poisson

2. Give the expected value and variance of X.

3. The cost of repairing a pothole is $ 5000. If Y denotes the county's pothole repair expense for next winter,find the mean value and variance of Y ?

2. Homework Equations and Attempt at a solution

1.) Pretty sure this is a Poisson random variable

2.) P =( \alphax * e -\alpha ) / x!

In this case α = 0.16 potholes/mile

x represents 0, 1, 2, ... , 30 is this correct?

Expected value of X= α = 0.16 potholes/mile
Variance of X = expected value of X = α = 0.16 potholes/mile

Y = aX + b

X = potholes that need to be fixed
a = 5000 (cost to fix each pothole)
b = 0


Expected value of Y = a * Expected value of X

Variance of Y = a2 * Variance of X
 
Physics news on Phys.org
tjackson said:
1. Homework Statement

During a typical Pennsylvania winter, I80 averages 1.6 potholes per 10 miles. A certain county is responsible for repairing potholes in a 30 mile stretch of the interstate. Let X denote the number of potholes the county will have to repair at the end of next winter.
1. The random variable X is

(i) binomial (ii) hypergeometric (iii) negative binomial (iv) Poisson

2. Give the expected value and variance of X.

3. The cost of repairing a pothole is $ 5000. If Y denotes the county's pothole repair expense for next winter,find the mean value and variance of Y ?

2. Homework Equations and Attempt at a solution

1.) Pretty sure this is a Poisson random variable

2.) P =( \alphax * e -\alpha ) / x!

In this case α = 0.16 potholes/mile

x represents 0, 1, 2, ... , 30 is this correct?
No. The random variable X denotes "the number of potholes the county will have to repair." Why would that number be limited to 30? ##\alpha## is the expected value of X, so it should be a number, not a number per mile.
 
You may be overcomplicating the problem a bit :)

Look at it this way: if, historically, the city averages about 1.6 potholes/ 10 miles, how many would you average in 30 miles?

Now to find the average expense of Y, there is a way we can look at it. Y = 5000*λ where lambda is the number of pot holes. So we can expect to pay say 5000*1.6 = $8000 for 10 miles of road. So how much would that be for 30 miles?

Now, let's talk about variance. Remember the definition of expected value and variance for Poisson? They are both λ .

So if you need to find Var(Y) = Var(5000*λ), what do we do with constant terms in variance? Hint: It's a large number, but that OK because variance isn't as helpful to know as standard deviation.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K