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.

Answer 2

The standard deviation of the situation is 1.88

The proportion of success is calculated as:

$p = \frac xn$

So, we have:

$p = \frac {68}{68 + 332}$

$p = 0.17$

The standard deviation is then calculated as:

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

For a shipment of a random sample of 25, we have:

$\sigma= \sqrt{25 * 0.17 * (1-0.17)}$

Evaluate the product

$\sigma= \sqrt{3.5275}$

Evaluate the root

$\sigma= 1.88$

Hence, the standard deviation of the situation is 1.88

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