Mathematics, 06.12.2019 03:31, meowmeowcow

Amachine is shut down for repairs if a random sample of 100 items selected from the daily output of the machine reveals at least 15% defectives. (assume that the daily output is a large number of items.) if on a given day the machine is producing only 10% defective items, what is the probability that it will be shut down? [hint: use the .5 continuity correction.]

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answered: bndkdiwjdjd

There is a 4.75% probability that it will be shut down.

Step-by-step explanation:

For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate 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:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .