- Hey guys I'm new to this. I'm not that big into probability but I

have to do some basic stuff as part of my computer degree. It's

giving me alot of trouble. Can any of you tell me the answer to these:

1.Distinguish between statistics of a sample and parameters of a

distribution.

2.An industrial process manufactures components with a 5% defective

rate. Each day products are inspected sequentially until the first

defective is obtained. Let X be the number of inspections carried out

until the first defective is found.

(a) If 5 products have been inspected without finding a defective,

what is the probability that the next 5 inspections will be free of

defectives?

(b) Derrive the expected number of inspections until the first

defective.

I know these questions might be easy to some of you guys but i'm

finding them difficult. - I looked at it this way. If Prob(x is defective) =

.05, then on average each observation has a 5%

defective part associated with it. Of course, we

don't look at it this way and we want to know when a

"whole" observation will be defective. So, when might

I get one whole defective unit? I said take 1 and

divide by .05 to get 20, so that on average, I should

expect to see a wholey defective unit on the twentieth

observation.

BTW, can anyone give me a counterexample where this

approach is not correct, or another problem of the

exact same nature where my back-of-the-envelope didn't

work? Remember, we aren't using conditional

probability here.

Regards,

M.W.

--- jason1990 <jason1990@...> wrote:> --- In probability@y..., Beata Stehlikova

__________________________________________________

> <beata_stehlikova@y...>

> wrote:

> > Intuitively I would say:

> > Probability 0.05 means that among 100 products

> there

> > will be on average 5 wrong. So among 20 products

> there

> > will be on average 1 wrong. I think you made your

> > equation in this way.

> >

> > But I don't know how to say that in this 'among 20

> we

> > expect 1' means 19 will be ok, and 20-th will be

> > wrong.

>

> I agree. It's not at all clear without a derivation.

> Intuitively, if

> one out of twenty will be defective, we ought to

> expect the defect to

> occur somewhere around the tenth observation.

>

> The fact that the "tail" (the very large waiting

> times with very

> small probability) is compensating things in just

> the right way so as

> to make the expected value 20 is not at all obvious.

>

> >

> > Also, how to show that this is what we really want

> -

> > expected value (which has its definition).

> >

> > Any comments on this are appreciated.

> >

> >

> > --- The Webmaster <maintainer_wiz@y...> wrote:

> > > Couldn't we have also simply said 1/.05 = 20,

> and

> > > gotten the answer this way?

> > >

> > > I loooked at the problem

> > > as such: 1/.05 = x/1, and then solved for x. I

> did

> > > appreciate your approach and maybe it holds for

> more

> > > complicated situations where a shortcut is not

> > > possible.

> > >

> > > Regards,

> > > M.W.

> > >

> > >

> > > --- Beata Stehlikova <beata_stehlikova@y...>

> > > wrote:

> > > > > 1.Distinguish between statistics of a sample

> and

> > > > > parameters of a

> > > > > distribution.

> > > >

> > > > Statistics of the sample is an estimate of the

> > > > parameter of distribution.

> > > >

> > > > >

> > > > > 2.An industrial process manufactures

> components

> > > > with

> > > > > a 5% defective

> > > > > rate. Each day products are inspected

> > > sequentially

> > > > > until the first

> > > > > defective is obtained. Let X be the number

> of

> > > > > inspections carried out

> > > > > until the first defective is found.

> > > > >

> > > > >

> > > > > (a) If 5 products have been inspected

> without

> > > > > finding a defective,

> > > > > what is the probability that the next 5

> > > > inspections

> > > > > will be free of

> > > > > defectives?

> > > > >

> > > >

> > > > I think that the production of defectives is

> > > > independent, so knowing about the quality of

> some

> > > > products doesn't say anything about the

> following.

> > > > So

> > > > the probability is (0.95)^5.

> > > >

> > > > > (b) Derrive the expected number of

> inspections

> > > > until

> > > > > the first

> > > > > defective.

> > > >

> > > >

> > > > (b) Derrive the expected number of inspections

> > > until

> > > > the first

> > > > defective.

> > > >

> > > > Number of inspections / probability

> > > > 1 / 0.05

> > > > 2 / 0.95 * 0.05

> > > > 3 / (0.95)^2 * 0.05

> > > > 4 / (0.95)^3 * 0.05

> > > > ...

> > > > i / (0.95)^(i-1) * 0.05

> > > >

> > > > Expected value =

> > > > 1 * 0.05 +

> > > > 2 * 0.95 * 0.05 +

> > > > 3 * (0.95)^2 * 0.05 +

> > > > ...

> > > > i * (0.95)^(i-1) * 0.05 +

> > > > ...

> > > >

> > > > = sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]

> > > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]

> > > >

> > > > Let's compute the sum

> > > > sum(i:1,infinity)[i*x^(i-1)] =

> > > > = sum(i:1,infinity)[d(x^i)/dx] =

> > > > = d[sum(i:1,infinity)(x^i)]/dx =

> > > > = d[x/1-x]/dx =

> > > > = 1/(1-x)^2

> > > >

> > > > So for x = 0.95 we have

> > > > sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2

> =

> > > 400

> > > > and so

> > > > expected value =

> > > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =

> > > > = 0.05 * 400 =

> > > > = 20

> > > >

> > > >

> > > >

> > > >

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