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# S2 June 14 - some issues? Watch

1. Hi - I just completed the June 14 S2 paper - I have a few small issues I wanted to get clarified, even if you're able to help out with just one I'd be much appreciative.

1 (b) - I got 1 mark for stating lambda = 3/2, I don't quite understand how the probability of the next patient arriving before 11:45 is = to 1-p(0), though? (Silly error/misunderstanding, I'm sure. Nonetheless I'd like to resolve it)

2 (d) - Surely this would be P(t>=7) / P(t>3), not p(t>7)/p(t>3)? Because it says 'at least 7 minutes' implying >=?

3 (d) - a good question to try for anyone with spare time! No help required for this one!

4 (c) - Again not really in need of help, but is there any clues that you'll need to flip the binomial from b(100,0.975) to b(100,0.025)?

6 (c) - Why is it that we set the 2/9 x - 1/9 bit = 0.5? Is it because the median is in between 1<x<4 (3rd value??)

Cheers, silly questions I am aware, but any help just to make sure I really get this module secure would be grately appreciated.

Zacken Ayman! etc, all the maths geniuses!
2. (Original post by iMacJack)
Hi - I just completed the June 14 S2 paper - I have a few small issues I wanted to get clarified, even if you're able to help out with just one I'd be much appreciative.

1 (b) - I got 1 mark for stating lambda = 3/2, I don't quite understand how the probability of the next patient arriving before 11:45 is = to 1-p(0), though? (Silly error/misunderstanding, I'm sure. Nonetheless I'd like to resolve it)

2 (d) - Surely this would be P(t>=7) / P(t>3), not p(t>7)/p(t>3)? Because it says 'at least 7 minutes' implying >=?

3 (d) - a good question to try for anyone with spare time! No help required for this one!

4 (c) - Again not really in need of help, but is there any clues that you'll need to flip the binomial from b(100,0.975) to b(100,0.025)?

6 (c) - Why is it that we set the 2/9 x - 1/9 bit = 0.5? Is it because the median is in between 1<x<4 (3rd value??)

Cheers, silly questions I am aware, but any help just to make sure I really get this module secure would be grately appreciated.

Zacken Ayman! etc, all the maths geniuses!
I sat this very paper. but it is too late in the day. If n one has got back to you later, you are welcome to quote me
3. (Original post by SeanFM)
I sat this very paper. but it is too late in the day. If n one has got back to you later, you are welcome to quote me
Bit of a hard paper!
4. (Original post by iMacJack)
1 (b) - I got 1 mark for stating lambda = 3/2, I don't quite understand how the probability of the next patient arriving before 11:45 is = to 1-p(0), though? (Silly error/misunderstanding, I'm sure. Nonetheless I'd like to resolve it)
They're calculating P(X >= 1). I'm not sure exactly why though, I'm struggling to understand the wording as well. Perhaps Zacken or Sean can clarify.

2 (d) - Surely this would be P(t>=7) / P(t>3), not p(t>7)/p(t>3)? Because it says 'at least 7 minutes' implying >=?
Continuous probability - whether your inequality is strict or not doesn't make a difference. You're correct in saying that it should be equal to from interpreting the language (and definitely would be correct for a discrete distribution), but in the calculation side of things it hardly makes a difference because a continuous random variable can't actually be equal to any one number; P(T = 7) = 0.

4 (c) - Again not really in need of help, but is there any clues that you'll need to flip the binomial from b(100,0.975) to b(100,0.025)?
Well, if you want to approximate using a Poisson distribution your lambda would be much too large if you approximated from B(100, 0.975).

6 (c) - Why is it that we set the 2/9 x - 1/9 bit = 0.5? Is it because the median is in between 1<x<4 (3rd value??)
There's no better way to do this than trial and error.

If you plug in the last value of the range into your first function, (x = 1 into x^2/9) you'll observe that F(1) = 1/9, which is much less than 0.5. If you plug in x = 4 into the second function, you'll get F(4) = 0.777..., which means that the median must lie here.
5. For Q1:

Switch the parameter to reflect the new time interval, hence 1.5.

Now, the probability that the next patient arrives in that time interval is the probability of X = 1 (that's one patient arrivingin the interval) or X = 2, or X = 3, etc...

So you want P(X >= 1) = 1 - P(X=0)
6. (Original post by Ayman!)
They're calculating P(X >= 1). I'm not sure exactly why though, I'm struggling to understand the wording as well. Perhaps Zacken or Sean can clarify.

Continuous probability - whether your inequality is strict or not doesn't make a difference. You're correct in saying that it should be equal to from interpreting the language (and definitely would be correct for a discrete distribution), but in the calculation side of things it hardly makes a difference because a continuous random variable can't actually be equal to any one number; P(T = 7) = 0.

Well, if you want to approximate using a Poisson distribution your lambda would be much too large if you approximated from B(100, 0.975).

There's no better way to do this than trial and error.

If you plug in the last value of the range into your first function, (x = 1 into x^2/9) you'll observe that F(1) = 1/9, which is much less than 0.5. If you plug in x = 4 into the second function, you'll get F(4) = 0.777..., which means that the median must lie here.
Ahhhhhhhhhh right I see! Thank you for the clarifications!

(Original post by Zacken)
For Q1:

Switch the parameter to reflect the new time interval, hence 1.5.

Now, the probability that the next patient arrives in that time interval is the probability of X = 1 (that's one patient arrivingin the interval) or X = 2, or X = 3, etc...

So you want P(X >= 1) = 1 - P(X=0)
So it's the probability of one OR more of the patients arriving, not only one? In which case, P(x>=1) makes sense.

Thank you!

Cheers guys!
7. (Original post by Zacken)
For Q1:

Switch the parameter to reflect the new time interval, hence 1.5.

Now, the probability that the next patient arrives in that time interval is the probability of X = 1 (that's one patient arrivingin the interval) or X = 2, or X = 3, etc...

So you want P(X >= 1) = 1 - P(X=0)
So any number of patients could arrive in the first 15 mins, so long that the first arrives then? Makes sense now.

(Original post by iMacJack)
Ahhhhhhhhhh right I see! Thank you for the clarifications!
No worries!
8. (Original post by Ayman!)
So any number of patients could arrive in the first 15 mins, so long that the first arrives then? Makes sense now.

No worries!
Yeah - I think the wording can be slightly misleading at times
9. (Original post by iMacJack)
Yeah - I think the wording can be slightly misleading at times
No, it's not misleading. There's a clear mathematical distinction between "one" and "exactly one"
10. (Original post by Zacken)
No, it's not misleading. There's a clear mathematical distinction between "one" and "exactly one"
Chill. I was talking about in general with S2.
11. (Original post by iMacJack)
Chill. I was talking about in general with S2.
I agree. My comprehension skills aren't very good.

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