Question

(1) A company manufacturer pipes which was sent out to customers in losts of 1000.The manufacturer operates a sampling scheme whereby a random sample 10 is taken from each lot ready for despatch and they are released only if the number of defective pipes in the sample is less than 3.Otherwise, the whole lot of 1000 is rejected and reprocessed using the Binomial p.d.f.
(I) if 5% of all the pipes produced are known to be defective, how many lots will be rejected out of 1000 lots processed?
(ii)If the producer replaces his entire pipe producing machines causing the number of defective pipes to drop to only 1% and releases lots if the number of defective pipes in the sample of 10 less than 2.How many lots per 1000 will he expect to save?

(b) The demand for a particular type pump at an isolated place is random and independent of previous occurrences, but the average demand in a week (7dsys) is for 2.8 pumps .Further supplies are ordered each Tuesday morning and arrive on weekly plan on Friday morning . Last Tuesday morning only one pumps was in the stock, so the stores man ordered six more to come on Friday morning .
(I) Find the probability that one pump will still be in stock on Friday morning when new stock arrives,
(ii) Find the probability that stock will be exhausted and there will be unsatisfied demand for at least one pump by Friday.
(iii) Find the probability that one pump will be in stock this Friday morning and at least five will be in stock next Tuesday morning.

Answer 1

Answer1: Total number of lots, N=1000, Sample taken ,n =10, so n<<N,

In this case , P would be the proportion of defective pipes, q would then be non-defective pipe.
out of N pipes.

Then we obtain a sampling distribution whose
mean μP and standard deviation σP are given by

Binomial distribution where, μ = p and σ = sqrt(pq)

It is found that 5% of total are defective so Mean, μ= .05

It is given that if more than 3 pipes are defective than lot will be rejected,

standard deviation = sqrt(p*q)/sqrt(n) = sqrt(.05*.95)/sqrt(10) = .069

Required probability = (area under normal curve to right of
z ), AS it is said that if greater than 3 pipes are defective ,then lot is rejected

z=(x-μ)/σ =(.3-.05)/.068 =3.6

by using the standard normal table we get required probability = 1-.4998 =.5

Hence out of 1000, 500 lots will be rejected.

ii. if no of defective pipe drops to 1%,and out of sample of 10 we need to have less than 2 pipes to be defective for lot to be rejected than,

standard deviation= sqrt(.01*.99)/sqrt(10)= .031

z=(x-μ)/σ =(.2-.01)/2*.031 =3.1

probability comes =1-.4990=.501 ,so out of 1000 he can save 501