Statistics, probability Help

[Calculator14]

Homework Statement

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.

Homework Equations

(work shown below)

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.

 

[tiny-tim]

Hi Calculator14!

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

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 ? ​

 

[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??

 

[tiny-tim]

he he!

yes ​

 

[DeeNos]

As a scientist, it is important to use statistics and probability to analyze and understand data. In this scenario, the number of defects in a sample of 20 parts is being observed for quality control purposes. Based on the data from the last year, we can calculate the expected number of defects in the next sample using probability.

Using the probabilities given (0 defects = 82%, 1 defect = 13%, 2 defects = 4%, and 3 defects = 1%), we can calculate the expected number of defects as follows:

Expected number of defects = (0 defects * 0.82) + (1 defect * 0.13) + (2 defects * 0.04) + (3 defects * 0.01)

= (0 * 0.82) + (1 * 0.13) + (2 * 0.04) + (3 * 0.01)

= 0 + 0.13 + 0.08 + 0.03

= 0.24

Therefore, we can expect to see an average of 0.24 defects in the next sample of 20 parts. It is important to note that this is an expected value and there is a possibility that the actual number of defects may differ from this value. However, using this calculation can help us make informed decisions and improve the quality control process.