Answer:
a) 44.93% probability that there are no surface flaws in an auto's interior
b) 0.03% probability that none of the 10 cars has any surface flaws
c) 0.44% probability that at most 1 car has any surface flaws
Step-by-step explanation:
To solve this question, we need to understand the Poisson and the binomial probability distributions.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.
So ![\mu = 10*0.08 = 0.8](https://tex.z-dn.net/?f=%5Cmu%20%3D%2010%2A0.08%20%3D%200.8)
(a) What is the probability that there are no surface flaws in an auto's interior?
Single car, so Poisson distribution. This is P(X = 0).
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.4493)
44.93% probability that there are no surface flaws in an auto's interior
(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?
For each car, there is a
probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.4493%29%5E%7B10%7D.%280.5507%29%5E%7B0%7D%20%3D%200.0003)
0.03% probability that none of the 10 cars has any surface flaws
(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?
At least 9 cars without surface flaws. So
![P(X \geq 9) = P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%209%29%20%3D%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.4493%29%5E%7B9%7D.%280.5507%29%5E%7B1%7D%20%3D%200.0041)
![P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.4493%29%5E%7B10%7D.%280.5507%29%5E%7B0%7D%20%3D%200.0003)
![P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%209%29%20%3D%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.0041%20%2B%200.0003%20%3D%200.0044)
0.44% probability that at most 1 car has any surface flaws