An automobile manufacturer is concerned about a fault in the braking mechanism of a particular model. The fault can, on rare occasions, cause a catastrophe at high speed. The distribution of the number of cars per year that will experience the catastrophe is a random variable with variance = 5. a) What is the probability that at most 3 cars per year will experience a catastrophe?
b) What is the probability that more than 1 car per year will experience a catastrophe?

(a) Probability that at most 3 cars per year will experience a catastrophe is 0.2650.

(b) Probability that more than 1 car per year will experience a catastrophe is 0.9596.

Step-by-step explanation:

We are given that the distribution of the number of cars per year that will experience the catastrophe is a Poisson random variable with variance = 5.

Let X = the number of cars per year that will experience the catastrophe

SO, X ~ Poisson()

The probability distribution for Poisson random variable is given by;

where, = Poisson parameter = 5  {because variance of Poisson distribution is only}

(a) Probability that at most 3 cars per year will experience a catastrophe is given by = P(X 3)

    P(X 3)  =  P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                   =  

                   =        

                   =  0.2650

(b) Probability that more than 1 car per year will experience a catastrophe is given by = P(X > 1)

               P(X > 1)  =  1 - P(X 1)

                             =  1 - P(X = 0) - P(X = 1)

                             =  

                             =  1 - 0.00674 - 0.03369

                             =  0.9596

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1