Question

a manufacturer has the following quality control check at the end of a production line. if at least 8 of 10 randomly picked articles meet all specifications, the whole shipment is approved. if in reality, 85% of a particular shipment meets all specifications, what is the probability that the shipment will make it through the control check?​

Answer 1

Using the binomial distribution, it is found that there is a 0.8202 = 82.02% probability that the shipment will make it through the control check.

For each article, there are only two possible outcomes, either it meets the specifications, or it does not. The probability of an article meeting the specifications is independent of any other article, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 10 articles are picked, hence $n = 10$.

• 85% of the articles meets all specifications, hence $p = 0.85$

The probability is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.85)^{8}.(0.15)^{2} = 0.2759$

$P(X = 9) = C_{10,9}.(0.85)^{9}.(0.15)^{1} = 0.3474$

$P(X = 10) = C_{10,10}.(0.85)^{10}.(0.15)^{0} = 0.1969$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2759 + 0.3474 + 0.1969 = 0.8202$

0.8202 = 82.02% probability that the shipment will make it through the control check.

For more on the binomial distribution, you can check brainly.com/question/24863377