Question

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?

Answer 1

Answer:

(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($\lambda = 5$)

The probability distribution for Poisson random variable is given by;

               $P(X=x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; \text{ where} \text{ x} = 0,1,2,3,...$

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

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

    P(X $\leq$ 3)  =  P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                   =  $\frac{e^{-5} \times 5^{0} }{0!} +\frac{e^{-5} \times 5^{1} }{1!} +\frac{e^{-5} \times 5^{2} }{2!} +\frac{e^{-5} \times 5^{3} }{3!}$

                   =  $e^{-5} +(e^{-5} \times 5) +\frac{e^{-5} \times 25 }{2} +\frac{e^{-5} \times 125}{6}$      

                   =  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 $\leq$ 1)

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

                             =  $1-\frac{e^{-5} \times 5^{0} }{0!} -\frac{e^{-5} \times 5^{1} }{1!}$

                             =  1 - 0.00674 - 0.03369

                             =  0.9596