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# Quick S2 Questions watch

1. 1. The time people take to complete a puzzle is modelled by continous uniform distribution 1<x<5 (less than or equal to and great than or equal to).

Given that someone has already spent 3minutes on the puzzle, find the probability that the person will complete the puzzle during the next minute.

2. p =0.2 n =50 Use poisson to find probability between 7 and 10 (inclusive)

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1. The answer is 0.159 but I seem to get 2/3. I know that (1/b-a) is 0.25. Can anyone remind me of the formula from S1 which is 'find the p given something else'?

2.The answer here is 0.4529 but I keep getting 0.3628 through P(X<10)-P(X<7)

The signs are less than or equal to btw. How is this method incorrect?
2. 1. p(a/b) = p(a and b)/p(b)

2. X-Po(10) p( x <or equal to 10) - p(x less than or equal to 6) < you need to to do 6 because you're including 7.
3. Thank you very much!
4. Sorry to double post as this is a new question:

Is there a formula to work out the IQR of a continuous uniform distribution?

For example find the IQR of X~U[100,150]? (The answer is 25)

Thanks in advance and will rep tomorow.
5. if i remember rightly, you don't need a formula. the lower quartile will just be 1/4 of the distribution, so 112.5.
then the upper quartile will be 3/4 of the distribution, so 137.5. and then 137.5 - 112.5 = 25.
it's all uniform so the quartiles are just nice and normal and even i think.

someone do correct me if i'm wrong.
6. Thanks again! I understand now
7. Sorry to bump someone else's topic, but where does the 0.159 answer come from in the first question you asked?
8. That is true.. I inserted in Brightgirl's equation but still cant seem to get 0.159...

Can anyone help? {will also +rep today}
9. I mean I would have said 0.5, just thinking about what can happen: essentially 2 minutes of the range have already happened, so really what it's asking is the probability of getting between 3 and 4 in a 3<x<5 distribution. So a half. But that's just a guess.

(Remember while they've already taken 3 minutes, everybody takes at least a minute, so that's irrevelent.)
10. Ahh I think you're probably correct. I could have just copied out the wrong answer, but can you explain your way of thinking in terms of using the distribution? I'm finding it hard picturing it your way..
11. Right okay.

You have an event P: say, normally, P(X<3). The probability of P is the ratio of the area which represents P to the area which represents all possible events (in terms of the graph). Now normally the latter, area which reps all events = 1. But in the case of something already having happened (conditional), some things are impossible.

In this question, originally we have a rectangle which starts at 1, ends at 5, height 0.25. So total area 1. But 2 minutes of this graph are now impossible, as the person has already taken 3 minutes (it starts at 1 which confuses matters). So really the rectangle of all events which could still happen is still height 0.25, but starts at 3, ends at 5: so area is now 0.5. Now question asks you what is chance of solution in next minute, i.e. between the 3rd and 4th minutes. Area of the rectangle which represents these events is 0.25*(4-3)=0.25.

Now, 0.25 is the probability of P(3<x<4) normally, because the ratio we are considering is 0.25:1. But in this case, the "possible" area is less, it's 0.5 as we said. So the ratio is 0.25:0.5 = 0.5: 1 = 0.5.

So that's where the probability of 0.5 comes from. Of all the events which could still occur, half of them are the ones we want.

Helps to draw a sketch, fiddle around with this as you read through above. It's rather hard to do in words but easy to see in terms of pictures. This approach works for any continuous distribution function really but it's easiest with uniform dists for obvious reasons (easy numbers, no integration).

Of course if it is 0.159, I'm baffled. Note this is basically a way of saying the same formula quoted above, but also a little sketch of why it works.
12. Thank you very much for your explanation.

So you worked out the second probability of the 0.25*(4-3) but then converted it back into the ratio of the original graph? Would it have been significant to know the ratio for the first probability of 'given he has already spent 1<x<3 minutes on the puzzle?

I'm going to try and re-read this slowly and draw out the graph again!
13. Ah, not quite. I worked out 0.25*(4-3): that IS the ratio with the original graph, as it were. But events aren't occurring in old graph they are occuring in a fraction of it, the 3 < 4 < 5 part. So that's what we compare it with. It does not state that he has already sepnt 1 < x < 3 minutes: he HAS, definitely, spent 3 minutes. There's no chance there.

Note that it doesn't matter in this question, all of the 'happening in next minute' occurs in the 'possible time for stuff to happen' part of graph (i.e. the former is completely contained in latter). This doesn't always happen in conditional probability: that's what the P(A and B) in the formula given above is about.

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