A shipment of 50,000 transistors arrives at a manufacturing plant. 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. What is the probability the engineer accepts the shipment

Respuesta :

Answer:

0.82% probability the engineer accepts the shipment

Step-by-step explanation:

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)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 500, p = 0.04[/tex]

So

[tex]E(X) = np = 500*0.04 = 20[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{500*0.04*0.96} = 4.3818[/tex]

What is the probability the engineer accepts the shipment?

Less than 10 defective. Using continuity correction, this is [tex]P(X < 10 - 0.5) = P(X < 9.5)[/tex], which is the pvalue of Z when X = 9.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{9.5 - 20}{4.3818}[/tex]

[tex]Z = -2.4[/tex]

[tex]Z = -2.4[/tex] has a pvalue of 0.0082

0.82% probability the engineer accepts the shipment