The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 airline passengers. In February,...

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The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 airline passengers. In February,...

The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 airline passengers. In February, XYZ airlines mishandled 4.05 bags per 1,000 passengers. Assume Poisson distribution.

What is the probability that in the next 1,000 passengers, XYZ will have no more than ten mishandled bags?
1. 0.982
2. 0.653
3. 0.997
4. 0.014

math Statistics-And-Probability

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Answer 1

Answer #1

Let X denotes the number of mishandles bags in a randomly selected group of 1000 passengers.

X ~ Poisson(4.05)

The probability mass function of X is

P(X = z) = e-40-4.05- =0.1.2.

The probability that in the next 1,000 passengers, XYZ will have no more than ten mishandled bags

10 10 , є_ 405 * 4.0 =0.996886 PlXS10) =ΣΡ(X = r) = Σ r!

ans-> c) 0.997

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Answer 2

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