Respuesta :
Answer:
There is a 0.62% probability that the engineer accepts the shipment.
The acceptance policy has a very low probability of being satisfied, so it is not sound.
Step-by-step explanation:
For each resistor, there are only two possible outcomes. Either it is defective, or it is not. This means that the binomial probability distribution will be used in our solution.
However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex].
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma,[/tex] the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X), \sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
The quality control engineer at the plant obtains a random sample of 500 resistors and will reject the entire shipment if 10 or more of the resistors are defective. Suppose that 4% of the resistors in the whole shipment are defective. This means that we have [tex]n = 500, p = 0.04[/tex].
What is the probability the engineer accepts the shipment?
This is the probability that there are at most 9 defective pieces. So this is the pvalue of Z when [tex]X = 9[/tex].
We have that:
[tex]\mu = E(X) = np = 500*0.04 = 20[/tex]
[tex]\sigma = \sqrt{V(x)} = \sqrt{np(1-p)} = \sqrt{500*0.04*0.96} = 4.38[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9 - 20}{4.38}[/tex]
[tex]Z = -2.51[/tex]
[tex]Z = -2.51[/tex] has a pvalue of 0.00621. This means that there is a 0.62% probability that the engineer accepts the shipment.
The acceptance policy has a very low probability of being satisfied, so it is not sound.
