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If an airline claims that it loses 5.5 bags a day from a busy airport, what is the probability that on the day you fly the airline will lose: 4 bags, Less than 3 8 or 9 bags.
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im curious if this is poisson based
\[P(x)=\frac{e^{-\mu}\mu^x}{x!}\]
a) 0.156 correct?
im unsure of the cdf, pretty sure of it, but cant say its correct. assuming it is then: e^(-5.5) (5.5)^4/4! = 0.1558 yes since there is nothing to indicate a normal distribution, this would be my best attempt
Yes, this is probably poisson based. It's homework stuff.
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yay!!
the rest is the addition of the required xs P(0) + P(1) + P(2) = P(x<3)
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