Mean and Standard Deviation of Binomial Distribution

Question

A company produces processing chips for cell phones. At one of its large factories, 2% of the chips produced are defective in some way. A quality check involves randomly selecting and testing 500 chips.

What are the mean and standard deviation of the number of defective chips in these samples?

Answer 1

Mean and Standard Deviation of Binomial Distribution

A company produces processing chips for cell phones. At one of its large factories, 2% of the chips produced are defective in some way. A quality check involves randomly selecting and testing 500 chips.

What are the mean and standard deviation of the number of defective chips in these samples?

Solution:

1.    Let X be the number of defective chips in any random sample of 500 chips.

2.    X will be a binomial variable. How do we know this?

a.    First, number of trials (500) is a finite number.

b.    Second, each trial is independent of any other trial.

c.    Third, the probability of number of defective chips across any random sample of 500 chips is the same and that probability is 2%.

d.    Hence, probability p = 2% = 0.02

3.    The formula for the mean of a binomial distribution is:

a.    Mean = mu(X) = E(X) = n*p = 500 * 0.02 = 10.

b.    This is logical as with 2% defect rate, we can expect 2 defects in any random sample of 100 chips and hence with 500 samples, the mean number of defective chips shall be 10.

c.    What this means is that, no matter, how we choose any random sample of 500 chips, 10 of them are expected to be defective (2%).

d.    Special Note: And further, 500 being a large number: n*p = 500*0.02 = 10>5 as well as n*(1-p) = 500*0.98 = 498 > 5. Hence, the binomial distribution in question can be approximated to a Standard Normal Distribution N(0,1).

4.    Now, the Standard Deviation of a binomial distribution:

a.    Standard Deviation: sigma(X) = sqrt(n*p*(1-p)) = sqrt(500*0.02*0.98) = sqrt(9.8) = 3.13 (rounded to 2 decimal places.

Answers:

Mean = 10

Standard Deviation: 3.13