Question

The number of surface flaws in a plastic roll used in the interior of automobiles has a Poisson distribution with a mean of 0.09 flaw per square foot of plastic roll. Assume an automobile interior contains 9 square feet of plastic roll. Round your answers to four decimal places (e.g. 98.7654).

(a) What is the probability that there are no surface flaws in an auto’s interior?
(b) If 21 cars are sold to a rental company, what is the probability that none of the 21 cars has any surface flaws?
(c) If 21 cars are sold to a rental company, what is the probability that at most 3 cars has any surface flaws?

Answer 1

Answer: a) 0.44, b) 0.00000033, b) 0.00101.

Step-by-step explanation:

Since we have given that

Mean = λ = 0.09 flaw per square foot of plastic

Number of square feet of plastic roll = 9

So, λ = 9 × 0.09 = 0.81

(a) What is the probability that there are no surface flaws in an auto’s interior?

P(X=0) is given by

$\dfrac{e^{-\lambda}\lambda^0}{0!}\\\\=e^{-0.81}\\\\=0.44$

(b) If 21 cars are sold to a rental company, what is the probability that none of the 21 cars has any surface flaws?

Here, n = 21

So, we will use "Binomial distribution":

p = 1-0.44=0.56

q = 0.44

P(X=21) is given by

$^{21}C_{21}(0.44)^{21}(0.56)^0\\\\=0.000000033$

(c) If 21 cars are sold to a rental company, what is the probability that at most 3 cars has any surface flaws?

$P(X\leq 3)=\sum _{x=0}^3 0.00101$

Hence, a) 0.44, b) 0.00000033, b) 0.00101.