Question

The average defect rate on a 2010 Volkswagen vehicle was reported to be 1.33 defects per vehicle. Suppose that we inspect 100 Volkswagen vehicles at random. (a) What is the approximate probability of finding at least 157 defects

Answer 1

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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\lambda$ 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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = n\lambda, \sigma = \sqrt{\lambda}\sqrt{n}$

The average defect rate on a 2010 Volkswagen vehicle was reported to be 1.33 defects per vehicle.

This means that $\lambda = 1.33$

Suppose that we inspect 100 Volkswagen vehicles at random.

This means that $n = 100$

Mean and standard deviation:

$\mu = n\lambda = 100*1.33 = 133$

$\sigma = \sqrt{\lambda}\sqrt{n} = \sqrt{1.33}\sqrt{100} = 11.53$

What is the approximate probability of finding at least 157 defects?

Using continuity correction(Poisson is a discrete distribution, normal continuous), this is $P(X \geq 157 - 0.5) = P(X \geq 156.5)$, which is 1 subtracted by the p-value of Z when X = 156.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{156.5 - 133}{11.53}$

$Z = 2.04$

$Z = 2.04$ has a p-value of 0.9793.

1 - 0.9793 = 0.0207

0.0207 = 2.07% approximate probability of finding at least 157 defects