Statistics, probability ! Help!

1. Jul 6, 2011

Calculator14

1. The problem statement, all variables and given/known data

The number of defects in a sample of 20 parts is recorded for quality control purposes; over the last year the number of defects and their occurrence rate has been : 0 defects, 82%; 1 defect, 13%; 2 defects, 4%; 3 defects, 1%. Find the expected number of defects in the next sample of 20 parts.

2. Relevant equations

(work shown below)

3. The attempt at a solution

1*.13+2*.04+3*.01=.24
20*.24=4.8 the expected number of defects in the next sample of 20 parts.

2. Jul 6, 2011

tiny-tim

Hi Calculator14!
hmm … you don't know when to stop, do you?

How can the expected number be between 4 and 5 when the given probabilities only go up to 3 ??

What do you think the 0.24 is ?

3. Jul 6, 2011

Calculator14

OHHHHHHHH!!! So I think I went a little too far with this, haha! My apologies tiny-tim, thank you for pointing out my mistake!! I believe my answer should be .24??

4. Jul 7, 2011

he he!

yes