Respuesta :
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]\mu = 0.02*10 = 0.2[/tex]
(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).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.2}*(0.2)^{0}}{(0)!} = 0.818731[/tex]
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 [tex]n = 10[/tex]
This probability is P(X = 10), that is, all the cars without flaws.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 10) = C_{10,10}.(0.818731)^{10}.(1-0.818731)^{0} = 0.135336[/tex]
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
[tex]P(X \geq 9) = P(X = 9) + P(X = 10)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{10,9}.(0.818731)^{9}.(1-0.818731)^{1} = 0.299631[/tex]
[tex]P(X = 10) = C_{10,10}.(0.818731)^{10}.(1-0.818731)^{0} = 0.135336[/tex]
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) = 0.299631 + 0.135336 = 0.434967[/tex]
43.4967% probability that at most 1 car has any surface flaws
Probability of an event is the measurement of its occurrence. The probability of the needed events is given by:
- P(There are no surface flaws in an auto's interior) = 0.819 approx
- P(None of the 10 cars has any surface flaws) = 0.136 approx
- If 10 cars are sold, the probability that at most 1 car has any surface flaws = 0.436 approx
How to find that a given condition can be modeled by binomial distribution?
Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
What are some of the properties of Poisson distribution?
Let X ~ Pois(λ)
Then we have:
E(X) = λ = Var(X)
Since standard deviation is square root (positive) of variance,
Thus,
Standard deviation of X = [tex]\sqrt{\lambda}[/tex]
Its probability function is given by
f(k; λ) = Pr(X = k) = [tex]\dfrac{\lambda^{k}e^{-\lambda}}{k!}[/tex]
For this case, let we take:
X = the number of surface flaws in plastic panels used in the interior of automobiles
Since 0.02 mean of flaw is per square foot and there is 10 sq. feet of plastic panel for each vehicle, thus,we get:
[tex]E(X) = \lambda = 0.02 \times 10 = 0.2\\\\X \sim \rm Pois(0.2)[/tex]
Thus, the probabilities for specified cases are evaluated as:
- Case 1: The probability that there are no surface flaws in an auto's interior
That means we've got X = 0, its probability is calculated using the probability function of Poisson distribution as:
[tex]P(X=k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}\\\\P(X = 0) = \dfrac{0.2^0 \times e^{-0.2}}{0!} \approx 0.819[/tex]
- Case 2: The probability that none of the 10 cars has any surface flaws
Each car having surface flaw is independent of other cars (assumingly), then we can use binomial distribution here:
Let Y = number of cars having no flaws.
P(Success = no flaw) = 0.819 (as derived in first case) = p
n = number of trials = 10
Thus, we get: [tex]Y \sim B(n=10, p = 0.819)[/tex]
Then, we get:
P(The probability that none of the 10 cars has any surface flaws) = P(all 10 cars have no surface flaws) = P(Y = 10)
Using the probability function of binomial distribution, we get:
[tex]P(Y=10) = \:^{10}C_{10}(0.819)^{10}(0.0.181)^{0} \approx 0.136[/tex]
- Case 3: If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws
Similar to above case, this probability is P(Y [tex]\geq[/tex] 9) (at least 9 success or no flaw cars, since this is equivalent to at most 1 car having surface flaw)
[tex]P(Y \geq 9) = P(Y =9) + P(Y = 10)\\\\P(Y \geq 9) = \: ^{10}C_9(0.819)^9(0.181)^1 + \: ^{10}C_{10}(0.819)^{10}(0.181)^0\\\\P(Y \geq 9) \approx 0.3 + 0.136 \approx 0.436[/tex]
Thus, the probability of the needed events is given by:
- P(There are no surface flaws in an auto's interior) = 0.819 approx
- P(None of the 10 cars has any surface flaws) = 0.136 approx
- If 10 cars are sold, the probability that at most 1 car has any surface flaws = 0.436 approx
Learn more about Poisson distribution here:
https://brainly.com/question/7879375
