A company produces a particular type of electrical components. The probability t

Question

A company produces a particular type of electrical components. The probability that a random component is defective is denoted by p (an unknown quantity). The individual components are defective / not defective independently. Every day one sample test is taken with 20 components controlled. Let X be the number of defective in such a sample test. a) Explain why X is bionomical distributed. Find E (X) and Var (X) expressed by p. b) The company accepts defective ratio of up to 0.05. To check the error ratio in the production, the following routine in the daily random sampling is done: If the number of defective in the sample test is 3 or more, the process stops so that it can be adjusted (because they now think that p can be greater than 0.05). What is the probability that the process will be stopped if p = 0.05? What is the probability that the process is stopped if p = 0.2? c) The components are packed in boxes with 10 components in each box. In a given box there are 3 defective components and 7 correct. You pull out 4 random without replacement. Let Y be the number of defective you get. Which probability distribution will Y get? What are the expectations of Y? Find the probability that you get exactly 1 defective by a draw.

Explanation / Answer

a) X is said to follow a binomial distribution because

i) the probability of component being defective is p remains unchanged for every component.

ii) The individual components are defective / not defective independently.

iii) there are 2 possible outcomes for each component: defective or non defective.

Every day one sample test is taken with 20 components controlled.  The probability that a random component is defective is denoted by p (an unknown quantity).

so X~Bin(20,p)

then E[X]=20p and variance=V[X]=20p(1-p)

b) if X>=3 then the process is stopped

when p=0.05

X~Bin(20,0.05)

now the probability that the process will be stopped is

P[X>=3]=1-P[X<3]=1-P[X=0]-P[X=1]-P[X=2]=1-20C00.0500.9520-20C10.0510.9519-20C20.0520.9518

=0.07548367 [answer]

now p=0.2

then the probability that the process will be stopped is

P[X>=3]=1-P[X<3]=1-P[X=0]-P[X=1]-P[X=2]=1-20C00.200.820-20C10.210.819-20C20.220.818

=0.7939153 [answer]

c) there are 10 components out of which 3 defective 7correct

4 things are drawn at random without replacement.

Y be the number of defectives

so Y can take the value 0,1,2,3

Y would follow a hypergeometric distribution with P[Y=y]=(3Cy*7C4-y)/10C4

probability of gettinge exactly 1 defective is P[Y=1]=3C1*7C3/10C4=0.5 [answer]

now P[Y=0]=7C4/10C4=0.16667

P[Y=1]=0.5

P[Y=2]=3C2*7C2/10C4=0.3

P[Y=3]=3C3*7C1/10C4=0.0333

so E[Y]=0*1.6667+1*0.5+2*0.3+3*0.0333=1.1999 [answer]

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