Question

he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the 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? (c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? Round your answers to four decimal places (e.g. 98.7654).

Answer 1

Answer:

a) 81.8731% probability that there are no surface flaws in an auto's interior.

b) 13.5336% probability that none of the 10 cars has any surface flaws

c) 43.4967% 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 distribution.

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson distribution with a mean of 0.02 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

This means that $\mu = 0.02*10 = 0.2$

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

Single vehicle, so we use the Poisson distribution.

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.2}*(0.2)^{0}}{(0)!} = 0.818731$

81.8731% 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?

81.87% probability that there are no surface flaws in an auto's interior. So p = 0.8187.

10 cars, so $n = 10$

This probability is P(X = 10), that is, all the cars without flaws.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.818731)^{10}.(1-0.818731)^{0} = 0.135336$

13.5336% 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 flaws, so

$P(X \geq 9) = P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.818731)^{9}.(1-0.818731)^{1} = 0.299631$

$P(X = 10) = C_{10,10}.(0.818731)^{10}.(1-0.818731)^{0} = 0.135336$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.299631 + 0.135336 = 0.434967$

43.4967% probability that at most 1 car has any surface flaws

Answer 2

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2  

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

$P(k)= \frac{0.2^ke^{-0.2}}{k!}$

a)

In this case, we are looking for P(0)

$P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%$

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and  

$P(k)= \frac{2^ke^{-2}}{k!}$

and  

$P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%$

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

$P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%$

Hence, the probability that at most one car has no flaws is 27.06%