Solution Found!
If ? > 0 and ? is a positive integer, the relationship
Chapter 4, Problem 99E(choose chapter or problem)
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.