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Normal approximation to Poisson random variable

875
17
1. Homework Statement
Suppose that the number of asbestos particles in a sam-
ple of 1 squared centimeter of dust is a Poisson random variable
with a mean of 1000. What is the probability that 10 squared cen-
timeters of dust contains more than 10,000 particles?

2. Homework Equations
[tex] E(aX+b) = aE(X) + b[/tex]
[tex]Var(aX) = a^2 Var(X) [/tex]

3. The Attempt at a Solution
Let X = number of asbestos particles in 1[itex]\mbox{cm}^2[/itex]. Define Y = number of asbestos particles in 10[itex]\mbox{ cm}^2[/itex]. So we have [itex]Y=10X[/itex]. Using the formula given above, we get [itex] E(Y)=10E(X)[/itex] and [itex]Var(Y) = 100 Var(X)[/itex]. But since X is a Poisson random variable, we have [itex]E(X) = \lambda = Var(X) = 1000[/itex]. So we get for Y variable, [itex]E(Y) = 10000[/itex] and [itex]Var(Y) = 100000[/itex]. Then the probability we need to find is [itex]P(Y > 10000)[/itex]. Now we use the Normal approximation here. [itex]E(Y) = 10000[/itex] and [itex]Var(Y) = 100000[/itex]. So [itex]P(Y \geq 10001.5)[/itex]. So we get the following expression
[tex]P\left(z \geq \frac{10001.5 - 10000}{\sqrt(100000)}\right)[/tex]
So now I use [itex]pnorm[/itex] function in [itex]R[/itex] , to calculate this probability. It is

[tex]\mbox{pnorm(10001.5, 10000, sqrt(100000), lower.tail=F)}[/tex]

which gives us [itex]0.4981077[/itex]. Is this right ? The solution manual for Montgomery and Runger says that [itex] E(Y) = \lambda = 10000 = Var(Y)[/itex]. Is that a mistake ?

thanks
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,720
1. Homework Statement
Suppose that the number of asbestos particles in a sam-
ple of 1 squared centimeter of dust is a Poisson random variable
with a mean of 1000. What is the probability that 10 squared cen-
timeters of dust contains more than 10,000 particles?

2. Homework Equations
[tex] E(aX+b) = aE(X) + b[/tex]
[tex]Var(aX) = a^2 Var(X) [/tex]

3. The Attempt at a Solution
Let X = number of asbestos particles in 1[itex]\mbox{cm}^2[/itex]. Define Y = number of asbestos particles in 10[itex]\mbox{ cm}^2[/itex]. So we have [itex]Y=10X[/itex]. Using the formula given above, we get [itex] E(Y)=10E(X)[/itex] and [itex]Var(Y) = 100 Var(X)[/itex]. But since X is a Poisson random variable, we have [itex]E(X) = \lambda = Var(X) = 1000[/itex]. So we get for Y variable, [itex]E(Y) = 10000[/itex] and [itex]Var(Y) = 100000[/itex]. Then the probability we need to find is [itex]P(Y > 10000)[/itex]. Now we use the Normal approximation here. [itex]E(Y) = 10000[/itex] and [itex]Var(Y) = 100000[/itex]. So [itex]P(Y \geq 10001.5)[/itex]. So we get the following expression
[tex]P\left(z \geq \frac{10001.5 - 10000}{\sqrt(100000)}\right)[/tex]
So now I use [itex]pnorm[/itex] function in [itex]R[/itex] , to calculate this probability. It is

[tex]\mbox{pnorm(10001.5, 10000, sqrt(100000), lower.tail=F)}[/tex]

which gives us [itex]0.4981077[/itex]. Is this right ? The solution manual for Montgomery and Runger says that [itex] E(Y) = \lambda = 10000 = Var(Y)[/itex]. Is that a mistake ?

thanks
E(Y) = Var(Y) = 10000 are correct, Whether or not those are equal to λ depends on how you define λ.

The figure 0.498 is plausible. Since the mean is so large, the Poisson distribution looks very much like the normal, and you are asking for values greater than the mean by 0.01 standard deviations or more (so the answer ought to be just a bit less than 1/2).
 
875
17
Hi Ray, But from the Y = 10X, we should have [itex]Var(Y) = 100 Var(X)[/itex].
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,720
Hi Ray, But from the Y = 10X, we should have [itex]Var(Y) = 100 Var(X)[/itex].
No. ##Y \neq 10X##. The ##10X## figure could be true only if each of the ten 1 cm2 pieces had exactly the same number of asbestos particles, so that 10 of them had exactly 10 times as many as the single one, and that would mean that there is no randomness at all. That does not describe your system.
 

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