Question

Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components . From the shipment you take a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part. The standard deviation of this situation is?

Answer 1

Answer:

The standard deviation for the number of defective parts in the sample is 1.88.

Step-by-step explanation:

The sample is with replacement, which means that the trials are independent, and thus, the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.  

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

68 defective out of 400:

This means that $p = \frac{68}{400} = 0.17$

From the shipment you take a random sample of 25.

This means that $n = 25$

Standard deviation for the number of defective parts in the sample:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{25*0.17*0.83} = 1.88$

The standard deviation for the number of defective parts in the sample is 1.88.