Answer:
Probability of more than 1 death in a corps in a year
12.52%
Probability of no deaths in a corps over 7 years
1.40%
Step-by-step explanation:
If the data collected follow a Poisson distribution with mean 0.61, then
Probability of k deaths in a year is
[tex]P(k\;deaths\;in\;one\;year)=\frac{0.61^ke^{-0.61}}{k!}[/tex]
The probability of more than 1 death in a corps in a year would be
(1)[tex]P(2)+P(3)+...=\sum_{k=2}^{\infty}P(k)=\sum_{k=2}^{\infty}\frac{0.61^ke^{-0.61}}{k!}=e^{-0.61}\sum_{k=2}^{\infty}\frac{0.61^k}{k!}[/tex]
But the Taylor series around 0 for the exponential function is
[tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex]
So,
[tex]e^{0.61}=\frac{0.61^0}{0!}+\frac{0.61^1}{1!}+\sum_{k=2}^{\infty}\frac{0.61^k}{k!}[/tex]
and
[tex]\sum_{k=2}^{\infty}\frac{0.61^k}{k!}=e^{0.61}-1.61[/tex]
Replacing in (1), we obtain
[tex]\sum_{k=2}^{\infty}P(k)=e^{-0.61}(e^{0.61}-1.61)=1-1.61e^{-0.61}=1-0.874794\approx 0.1252[/tex]
or 12.52%
b)
The probability of 0 deaths in one year is
[tex]P(0)=e^{-0.61}[/tex]
As the events are independent, the probability of 0 deaths over 7 years is the product of P(0) by itself 7 times:
[tex]P(0))^7=(e^{-0.61})^7=e^{-4.27}\approx 0.0140[/tex]
or 1.40%
