Calculating Variance of a Sum of Independent Random Variables

  • Thread starter Gauss M.D.
  • Start date
  • Tags
    Poisson
In summary, during a two hour window, people are given the option of calling number X, donating $9.90, or number Y, donating $0.50. X is Poisson distributed with 1500 calls/minute and Y is Poisson distributed with 3750 calls/minute. To find the probability that more than $2,000,000 is raised, we let Z = 9.9X + 0.5Y and use the formula for the variance of Z in terms of the variances of X and Y. This formula can be used to approximate Z as a normal distribution with mean 2,007,000.
  • #1
Gauss M.D.
153
1

Homework Statement



During a two hour window, people are given the option of calling number X, donating $9.90, or number Y, donating $0.50.

X is Poisson distributed with 1500 calls/minute. Y is Poisson with 3750 calls/minute.

What is the probability that more than $2,000,000 is raised?

Homework Equations





The Attempt at a Solution



X = number of calls to number X in 120 minutes = Po(120*1500) = Po(180000)

Y = number of calls to number Y in 120 minutes = Po(120*3750) = Po(450000)

Let Z = 9.9X + 0.5Y. We're looking for P(Z > 2,000,000).

Z should have an expectation of 1,782,000 + 225,000 = 2,007,000.

Now I want to approximate Z = Po(μ) with Z ≈ N(μ,σ). But I can't work out the standard deviation for Z. Using [itex]\sqrt{μ}[/itex] doesn't give me the right answer.
 
Physics news on Phys.org
  • #2
Gauss M.D. said:

Homework Statement



During a two hour window, people are given the option of calling number X, donating $9.90, or number Y, donating $0.50.

X is Poisson distributed with 1500 calls/minute. Y is Poisson with 3750 calls/minute.

What is the probability that more than $2,000,000 is raised?

Homework Equations





The Attempt at a Solution



X = number of calls to number X in 120 minutes = Po(120*1500) = Po(180000)

Y = number of calls to number Y in 120 minutes = Po(120*3750) = Po(450000)

Let Z = 9.9X + 0.5Y. We're looking for P(Z > 2,000,000).

Z should have an expectation of 1,782,000 + 225,000 = 2,007,000.

Now I want to approximate Z = Po(μ) with Z ≈ N(μ,σ). But I can't work out the standard deviation for Z. Using [itex]\sqrt{μ}[/itex] doesn't give me the right answer.

I would bet that you have already seen how to do it, but may have forgotten. So, if X and Y are independent random variables and a, b are numbers, how does the variance of Z = a*X + b*Y relate to Var(X) and Var(Y)? There is a standard formula. It is used over and over and over again, so you should get to know it if you don't already or have forgotten it.
 

Related to Calculating Variance of a Sum of Independent Random Variables

1. What is the Poisson distribution?

The Poisson distribution is a probability distribution used to model the number of events that occur in a fixed interval of time or space, given the average rate at which these events occur.

2. What is the formula for approximating the Poisson distribution?

The formula for approximating the Poisson distribution is P(x;λ) = (e^(-λ) * λ^x) / x!, where x is the number of events and λ is the average rate of events.

3. When is the Poisson distribution used?

The Poisson distribution is commonly used in situations where the number of events occurring in a fixed interval of time or space is rare and independent of each other. This includes areas such as insurance, finance, and telecommunications.

4. What are the assumptions of the Poisson distribution?

The main assumptions of the Poisson distribution are that the events occur independently of each other, the average rate of events is constant, and the events cannot occur simultaneously.

5. How is the Poisson distribution different from other probability distributions?

The Poisson distribution is unique in that it only has one parameter (λ) and assumes that events occur independently and at a constant rate. Other distributions, such as the binomial and normal distributions, have more parameters and make different assumptions about the data. Additionally, the Poisson distribution is used for rare events, while other distributions may be used for more common events.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
893
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
5K
Back
Top