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Sagot :
Answer:
Hence the answers are,
a) Probability that in the past 12 months the car had more than one problem = P(X > 1) is 0.2716.
b) The Probability that in the past 12 months the car had almost two problems = P( X < 2) is 0.9160.
c) The Probability that in the past 12 months the car had zero problems = P(X= 0 ) is 0.3606.
Step-by-step explanation:
Let's take X to be the number of problems per car.
By considering the given statement, X follows a Poisson Distribution with Mean (X) = 1.02.
The Poisson probability formula is :
e Pr( X = k) = e- k! k= 0,1,2...
a)
The Probability that in the past 12 months the car had more than one problem = P(X > 1)
[tex]P(X > 1) =1- P(X < 1) \\\\=1- (P(X = 0) + P(X = 1)-1.021.02 e + 1.021.02 =1-6 0!\\= 1-0.3606 + 0.3678\\= 1-0.7284\\= 0.2716[/tex]
b)
The Probability that in the past 12 months the car had almost two problems = PX < 2)
[tex]Pr(X < 2) = Pr(X = i) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)\\-1.021.020 -1.021.02 -1.021.02 e e + e + 0! 1! 2!\\= 0.3606 + 0.3678 + 0.1876\\= 0.9160[/tex]
c)
The Probability that in the past 12 months the car had zero problems = P(X= 0 )
[tex]- 1.021.02 e 0!\\= 0.3606[/tex]
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