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**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