Poisson Distribution flaws in materials

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Suppose the number of flaws in a certain type of material can be modeled as a Poisson random variable. If flaws appear on average once in every $150$ square meters, what is the probability of finding at most one flaw in $225$ square meters? State your answer to three decimal places.

If $X$ is distributed as a Poisson

$Pr[X=x]= e^-lambda lambda^x/x!$

$E(X)=1.5$

$f(x)=1.5^x e^-1.5/x!$

$Pr(x=0)=e^-1.5 1.5^0/0!= 0,223$

$Pr(X<1) = 1-Pr(X = 0) = 1-0,223=0,777$

Is my procedure correct?

edited Sep 17 '17 at 19:59

Michael Hardy

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asked Sep 2 '17 at 6:45

marco lecci

573

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up vote
1
down vote

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If $X$ is distributed as a Poisson

$Pr[X=x]= e^-lambda lambda^x/x!$

$E(X)=1.5$

$f(x)=1.5^x e^-1.5/x!$

$Pr(x=0)=e^-1.5 1.5^0/0!= 0,223$

$Pr(X<1) = 1-Pr(X = 0) = 1-0,223=0,777$

Is my procedure correct?

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

asked Sep 2 '17 at 6:45

marco lecci

573

add a commentÂ |Â

up vote
1
down vote

favorite

If $X$ is distributed as a Poisson

$Pr[X=x]= e^-lambda lambda^x/x!$

$E(X)=1.5$

$f(x)=1.5^x e^-1.5/x!$

$Pr(x=0)=e^-1.5 1.5^0/0!= 0,223$

$Pr(X<1) = 1-Pr(X = 0) = 1-0,223=0,777$

Is my procedure correct?

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

asked Sep 2 '17 at 6:45

marco lecci

573

If $X$ is distributed as a Poisson

$Pr[X=x]= e^-lambda lambda^x/x!$

$E(X)=1.5$

$f(x)=1.5^x e^-1.5/x!$

$Pr(x=0)=e^-1.5 1.5^0/0!= 0,223$

$Pr(X<1) = 1-Pr(X = 0) = 1-0,223=0,777$

Is my procedure correct?

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

asked Sep 2 '17 at 6:45

marco lecci

573

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

edited Sep 17 '17 at 19:59

Michael Hardy

204k23187463

asked Sep 2 '17 at 6:45

marco lecci

573

asked Sep 2 '17 at 6:45

marco lecci

573

asked Sep 2 '17 at 6:45

marco lecci

573

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Your mean and Poisson distribution formula are correct but "at most one flaw" is same as $P(X le 1)$ and not $P(X<1)$.

Also, $P(X<1) = P(X=0) $ ; $P(Xge 1) = 1 - P(X<1) = 1- P(X=0)$

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

add a commentÂ |Â

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Your mean and Poisson distribution formula are correct but "at most one flaw" is same as $P(X le 1)$ and not $P(X<1)$.

Also, $P(X<1) = P(X=0) $ ; $P(Xge 1) = 1 - P(X<1) = 1- P(X=0)$

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

add a commentÂ |Â

up vote
2
down vote

accepted

Your mean and Poisson distribution formula are correct but "at most one flaw" is same as $P(X le 1)$ and not $P(X<1)$.

Also, $P(X<1) = P(X=0) $ ; $P(Xge 1) = 1 - P(X<1) = 1- P(X=0)$

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

add a commentÂ |Â

up vote
2
down vote

accepted

Your mean and Poisson distribution formula are correct but "at most one flaw" is same as $P(X le 1)$ and not $P(X<1)$.

Also, $P(X<1) = P(X=0) $ ; $P(Xge 1) = 1 - P(X<1) = 1- P(X=0)$

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

Your mean and Poisson distribution formula are correct but "at most one flaw" is same as $P(X le 1)$ and not $P(X<1)$.

Also, $P(X<1) = P(X=0) $ ; $P(Xge 1) = 1 - P(X<1) = 1- P(X=0)$

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

edited Aug 10 at 15:48

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

answered Sep 2 '17 at 9:38

BRAINSTELLAR

363

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

add a commentÂ |Â

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

Thank you for your help
â€“Â marco lecci
Sep 2 '17 at 22:16

add a commentÂ |Â

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