Respuesta :
Answer:
0.0207 = 2.07% approximate probability of finding at least 157 defects
Step-by-step explanation:
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^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\lambda[/tex] is the mean in the given interval.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
n instances of a Poisson distribution can be approximated to a normal distribution, with [tex]\mu = n\lambda, \sigma = \sqrt{\lambda}\sqrt{n}[/tex]
The average defect rate on a 2010 Volkswagen vehicle was reported to be 1.33 defects per vehicle.
This means that [tex]\lambda = 1.33[/tex]
Suppose that we inspect 100 Volkswagen vehicles at random.
This means that [tex]n = 100[/tex]
Mean and standard deviation:
[tex]\mu = n\lambda = 100*1.33 = 133[/tex]
[tex]\sigma = \sqrt{\lambda}\sqrt{n} = \sqrt{1.33}\sqrt{100} = 11.53[/tex]
What is the approximate probability of finding at least 157 defects?
Using continuity correction(Poisson is a discrete distribution, normal continuous), this is [tex]P(X \geq 157 - 0.5) = P(X \geq 156.5)[/tex], which is 1 subtracted by the p-value of Z when X = 156.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{156.5 - 133}{11.53}[/tex]
[tex]Z = 2.04[/tex]
[tex]Z = 2.04[/tex] has a p-value of 0.9793.
1 - 0.9793 = 0.0207
0.0207 = 2.07% approximate probability of finding at least 157 defects
