If ? > 0 and ? is a positive integer, the relationship

Chapter 4, Problem 99E

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QUESTION:

If \(\lambda>0\) and \(\alpha\) is a positive integer, the relationship between incomplete gamma integrals and sums of Poisson probabilities is given by

                      \(\frac{1}{\Gamma(\alpha)} \int_{\lambda}^{\infty} y^{\alpha-1} e^{-y}\) \(d y=\sum_{x=0}^{\alpha-1} \frac{\lambda^{x} e^{-\lambda}}{x !}\).

a If  has a gamma distribution with \(\alpha=2\) and \(\beta=1\), find \(P(Y>1)\) by using the preceding equality and Table 3 of Appendix 3 .
b
Applet Exercise If  has a gamma distribution with \(\alpha=2\) and \(\beta=1\), find \(P(Y>1)\) by using the applet Gamma Probabilities.

Equation Transcription:

Text Transcription:

lambda>0

alpha

1 over Gamma(alpha) integral y to infinity y^alpha-1 e^-y

dy=sum x=0 alpha-1 lambda^x e^-lambda over x!

alpha=2

beta=1

P(Y>1)

alpha=2

beta=1

P(Y>1)

Questions & Answers

QUESTION:

If \(\lambda>0\) and \(\alpha\) is a positive integer, the relationship between incomplete gamma integrals and sums of Poisson probabilities is given by

                      \(\frac{1}{\Gamma(\alpha)} \int_{\lambda}^{\infty} y^{\alpha-1} e^{-y}\) \(d y=\sum_{x=0}^{\alpha-1} \frac{\lambda^{x} e^{-\lambda}}{x !}\).

a If  has a gamma distribution with \(\alpha=2\) and \(\beta=1\), find \(P(Y>1)\) by using the preceding equality and Table 3 of Appendix 3 .
b
Applet Exercise If  has a gamma distribution with \(\alpha=2\) and \(\beta=1\), find \(P(Y>1)\) by using the applet Gamma Probabilities.

Equation Transcription:

Text Transcription:

lambda>0

alpha

1 over Gamma(alpha) integral y to infinity y^alpha-1 e^-y

dy=sum x=0 alpha-1 lambda^x e^-lambda over x!

alpha=2

beta=1

P(Y>1)

alpha=2

beta=1

P(Y>1)

ANSWER:

Step1 of 4: Goal

a

If Y has a gamma distribution with α = 2 and β = 1, find P(Y > 1) by using for  and Table 3 of Appendix 3.

 b  

If Y has a gamma distribution with α = 2 and β = 1, find P(Y > 1) by using the applet Gamma Probabilities.

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