Poisson Distribution

A cement pylon is considered safe if the hairline cracks are spread out and occur on the average of 1.2 cracks per 10-ft length of cement. In a 40-ft section of the cement pylon, what's the probability of 4 or more hairline cracks?

$\displaystyle P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}$

So $\displaystyle \lambda$ is the average rate (1.2 per 40ft?) and $\displaystyle x$ is the # of events.

?

Quote:

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Originally Posted by JJ007

A cement pylon is considered safe if the hairline cracks are spread out and occur on the average of 1.2 cracks per 10-ft length of cement. In a 40-ft section of the cement pylon, what's the probability of 4 or more hairline cracks?

$\displaystyle P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}$

So $\displaystyle \lambda$ is the average rate (1.2 per 40ft?) and $\displaystyle x$ is the # of events.

?

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I would think that $\displaystyle \lambda=4.8$ cracks per 40 ft.

Thus, $\displaystyle \mathbb{P}(X\geq 4) = 1-\mathbb{P}(X\leq 3)=1-\displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!}\approx 0.7058$

Does this make sense?

Quote:

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Originally Posted by Chris L T521

I would think that $\displaystyle \lambda=4.8$ cracks per 40 ft.

Thus, $\displaystyle \mathbb{P}(X\geq 4) = 1-\mathbb{P}(X\leq 3)=1-\displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!}\approx 0.7058$

Does this make sense?

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How did you get 4.8? Also did you use 3! for x! ?

Thanks

You have written 1.2 cracks per 10 feet. SO for 40 feet, you would have 1.2*4=4.8 cracks.

And, for your 2nd question, Chris L T521's notation given above means:

$\displaystyle \displaystyle\sum\limits_{x=0}^3\dfrac{(4.8)^xe^{-4.8}}{x!} = \frac{(4.8)^0 e^{-4.8}}{0!}+\frac{(4.8)^1 e^{-4.8}}{1!}+\frac{(4.8)^2 e^{-4.8}}{2!}+\frac{(4.8)^3 e^{-4.8}}{3!}$