Answer:
a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.
Step-by-step explanation:
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:
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.
To use the Poisson approximation for the binomial, we have that:
![\mu = np](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np)
1 in every 2500 automobiles produced has a particular manufacturing defect.
This means that ![p = \frac{1}{2500} = 0.0004](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B2500%7D%20%3D%200.0004)
a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.
This is
when
. So
![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 = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B7000%2C4%7D.%280.0004%29%5E%7B4%7D.%280.9996%29%5E%7B6996%7D%20%3D%200.1558)
0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.
Using the approximation:
. So
![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 = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-2.8%7D%2A%282.8%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.1557)
0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.