I could use some help with the problem below If you could pr

I could use some help with the problem below. If you could provide some explanation along with the answer that would be of great help since I am sure I will have this on the upcoming exam.

The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eleven syringes taken from the batch. Suppose the batch contains 5% defective syringes.

Below is a histogram showing the probabilities of r = 0, 1, 2, 3, ..., 10 and 11 defective syringes in a random sample of eleven syringes.

(a) Find . (Enter your answer to two decimal places.)
= syringes

What is the expected number of defective syringes the inspector will find? (Enter your answer to two decimal places.)
syringes

(b) What is the probability that the batch will be accepted? (Round your answer to three decimal places.)


(c) Find . (Round your answer to three decimal places.)
= syringes

Solution

see....Think from the perspective of the quality-control inspector...your goal is - search for defective syringes from the batch...
So, define \"success\" = If you have found a defective syringe

and \"failure\" = if you have found a good syringe...okay?

Total syringes in the batch = 11.....prob. of getting 1 defective syringe = 0.05..i.e, prob of success = 0.05..
So, this is a Binomial distributin with n= 11 and p=0.05...

Now,   = expected number of defective syringes the inspector will find = mean of this binomial distribution with n=11 and p=0.05 = n* p = 0.55....

(b) see...when a batch will be accepted?....when there are less than 2 defective syringes in the batch..right?...because,if the defective items becomes >= 2 , you have to reject the batch!....Now, when a batch have less than 2 defective syringes...what are the possibilities?...you can either have 1 defective syringe or you can have 0 defective syringes...right??

so,prob(batch will be accepted) = prob ( there are less than 2 defective syringes in the batch)
= prob( defective syringe < 2 ) = prob ( defective syringe = 1 ) + prob ( defective syringe = 0)

= ( 11 c 1 ) * ( 0.05)^1 * ( 1 - 0.05) ^10 + (11 c 0) * ( 0.05)^0 * ( 0.95)^11

= 0.3293+ 0.5688 = 0.8981... (cross check= see in the histogram you have posted , when the defective item = 0 , the prob. is more than 0.5 and when defective item = 1 , the prob. is more than 0.3....so, we have got the right answer!hurrah)

(c) = standard deviation of the binomial distribution with n=11 and p= 0.05...= sqrt ( np(1-p) ) = sqrt( 11*0.05*0.95) = 0.723

and do give a great exam :D