Thread: Reasonably likely 2nd

At a large publishing company 31 percent of the workers are proof readers. 16 employees are randomly selected for a special project.

What would be the reasonably likely numbers of proof readers in this sample of 16 employees?

Again, i tried the z score 1.96 but it seems to be wrong.

all in all, i had 1.3561 and 8.5639 as solutions

Originally Posted by makaveli89

At a large publishing company 31 percent of the workers are proof readers. 16 employees are randomly selected for a special project.

What would be the reasonably likely numbers of proof readers in this sample of 16 employees?

Again, i tried the z score 1.96 but it seems to be wrong.

all in all, i had 1.3561 and 8.5639 as solutions

Let X be the random variable number of proof readers.

"large" means you can use the binomial distribution.

X ~ Binomial(n = 16, p = 0.31).

E(X) = .....

Alternatively, you might prefer to find the value of x that makes Pr(X = x) largest.