Answer: a) 0.0481, b) 0.921, c) 13
Step-by-step explanation:
Since we have given that
[tex]\mu=8[/tex]
We will use Poisson distribution:
a) What is the probability that 12 fatalities occur in a specific week?
[tex]P(X=x)=\dfrac{e^{-\lambda}\times \lambda^x}{x!}\\\\P(X=12)=\dfrac{e^{-8}\times 8^{12}}{12!}\\\\P(X=12)=0.0481[/tex]
b) What is the probability that at least 12 fatalities occur in a specific week?
[tex]P(X\leq 12)=P(\dfrac{X-\lambda}{\sqrt{\lambda}}\leq \dfrac{12-\lambda}{\sqrt{\lambda}})\\\\P(X\leq 12)=P(z\leq \dfrac{12-8}{\sqrt{8}})\\\\P(X\leq 12)=P(z\leq 1.4142)\\\\P(X\leq 12)=0.921[/tex]
c) How many motor-vehicle fatalities would have to occur during a given week to conclude there are an unusually high number of events in that week?
[tex]\dfrac{x_0-\lambda}{\sqrt{\lambda}}=1.645\\\\\dfrac{x_0-8}{\sqrt{8}}=1.645\\\\x_0-8=1.645\times \sqrt{8}=4.652\\\\x=4.652+8\\\\x=12.652\\\\x\approx 13[/tex]
Hence, a) 0.0481, b) 0.921, c) 13
