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# S2 distribution question watch

1. 2. A manufacturer produces computer monitor screens in which the picture is
composed of 780 000 pixels. On average 1 in 500 000 of these pixels is faulty.
Let X represent the number of faulty pixels in one monitor screen.

A retailer orders a batch of 5 monitor screens from the manufacturer. (The batch may be regarded as a random sample)

iv) The manufacturer wishes to improve quality so that 90% of the monitor screens have no faulty pixels at all. To what value must the manufacturer reduce the probability of a piexl being faulty in order to achieve this?

I don't understand what this question is asking, is it 90% of the screens in the batch that he wants to reduce the probability of?

I tried doing the following:
X~Po(1.56)
P(X=0) = 0.21

90% of 5 = 4.5 so 4.5 screens must have no faulty pixels
(0.210^4.5) * (0.790^0.5) = 7.9x10^-4

However this doesnt seem right, since now the probability that there can be a faulty pixel per screen is bigger than before? where have i gone wrong?
2. (Original post by IDontKnowReally)
However this doesnt seem right, since now the probability that there can be a faulty pixel per screen is bigger than before? where have i gone wrong?
You can't have 4.5 screens, it has to be an integer. So, the 90% cannot be refering to the batch of 5.

If you need further, I'd need to see the whole question, link/photo.
3. (Original post by ghostwalker)
You can't have 4.5 screens, it has to be an integer. So, the 90% cannot be refering to the batch of 5.

If you need further, I'd need to see the whole question, link/photo.
I'm not too sure how to upload a photo but here is the question as it was given to me:

A manufacturer produces computer monitor screens in which the picture is composed of 780 000 pixles. On average 1 in 500 000 of these pixels is faulty. Let X represent the number of faulty pixels in one monitor screen.
i) State the condition required for X to be binomially distributed

ii) Explain why a Poisson distribution provides a good approximation for X. State the mean of this Poisson distribution

iii) Hence find the probability that a randomly selected monitor screen has
(A) exactly one fauulty pixel
(B) at least two faulty pixels

iv) A retailer orders a batch of five monitor screens from the manufacturer. (The batch may be regarded as a random sample.)
(A) Find the probability that exactly one of the five monitor screens has at least two faulty pixels
(B) Find the probabilty that there are at most 10 faulty pixels n total in the batch

v) The manufacturer wishes to improve the quality so that 90% of the monitor screens have no faulty pixels at all. To what value must the manufacturer reduce the probablity of a pixel being faulty in order to achieve this?
4. (Original post by IDontKnowReally)
v) The manufacturer wishes to improve the quality so that 90% of the monitor screens have no faulty pixels at all. To what value must the manufacturer reduce the probablity of a pixel being faulty in order to achieve this?
OK. The batch of 5, seems to refer solely to part (iv) of the question.

Part (v) is separate. So you're looking to find the probability that a pixel is faulty, which gives rise to the probability of no faults on a screen equalling 0.9
5. (Original post by ghostwalker)
OK. The batch of 5, seems to refer solely to part (iv) of the question.

Part (v) is separate. So you're looking to find the probability that a pixel is faulty, which gives rise to the probability of no faults on a screen equalling 0.9
so that just means the probability that a pixel is faulty is 0.1?
6. (Original post by IDontKnowReally)
so that just means the probability that a pixel is faulty is 0.1?
I'm pretty sure you meant "probability that a screen is faulty is 0.1".

Yep - that's it.
7. (Original post by ghostwalker)
I'm pretty sure you meant "probability that a screen is faulty is 0.1".

Yep - that's it.
Sorry, yeah, that's what I meant. But how do I go about answering the question?
8. (Original post by IDontKnowReally)
Sorry, yeah, that's what I meant. But how do I go about answering the question?
Well you've already gone through the process of working out the probability that a screen doesn't fail (no faults), for a given pixel failure probability. So:

Assume probability of pixel failure is "p". Work out the prob. that the screen doesn't fail in terms of p. Equate to 0.1 and solve for p.
9. (Original post by ghostwalker)
Well you've already gone through the process of working out the probability that a screen doesn't fail (no faults), for a given pixel failure probability. So:

Assume probability of pixel failure is "p". Work out the prob. that the screen doesn't fail in terms of p. Equate to 0.1 and solve for p.
Got it, thank you! Didn't quite realise that it was referring to the probability of faulty pixels on one monitor screen. But why is it 0.1 and not 0.9 that you equate to?
10. (Original post by IDontKnowReally)
Got it, thank you! Didn't quite realise that it was referring to the probability of faulty pixels on one monitor screen. But why is it 0.1 and not 0.9 that you equate to?
My mistake - should be 0.9, the probaility of no faults.

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