The number of defects in a random sample of 200 parts produced by a machine is binomially distributed with p = .03. Based on this information, the standard deviation of the number of defects in the sample is 5.82.

Question

Quizplus · Accepted Answer

The standard deviation of a binomial distribution with parameters n and p is:

sigma = sqrt(np(1-p))

Using the given information, we can solve for n:

n = 200 
p = 0.03 
sigma = 5.82

5.82 = sqrt(200 * 0.03 * (1-0.03)) = 5.82

Therefore, the statement is true.

The Number of Defects in a Random Sample of 200