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# Edexcel S2 - 27th June 2016 AM watch

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1. (Original post by M3WIZARD)
Any thoughts on tomorrow's paper? - Hopefully it's going to be a 6 Question paper ending on a nice easy mean or median sampling distribution
Impossible to know
2. (Original post by RetroSpectro)
The only Approximation that can be used for a Poisson distribution is a normal approx.
Yeah but how do you know you have to use normal approx. it doesnt say you use it
3. I just found an interesting question (January 2002).It says the bus arrives between 5 minutes early and 9 minutes late.
Therefore, it defined the distribution as X~U[-5,9].However, the markscheme defines it as X~U[0,14]. As a result, I got everything wrong, except part d.Do you think I would've received the marks anyway? How would you proceed with this question?

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4. (Original post by fpmaniac)
Yeah but how do you know you have to use normal approx. it doesnt say you use it
It says 'using a suitable approximation'
5. How I remember skewness is by writing the values in the order MEAN MEDIA AND MODE
if mean<median<mode, it is negatively skewed because the pointy part of the inequality is pointing in the negative direction (think of a number line, the negative values are to the left of zero) and if MEAN>MEDIAN>MODE, then it is positively skewed as the pointy parts points in the positive direction. (Just my own odd way of remembering it
6. (Original post by apzoe)
I just found an interesting question (January 2002).It says the bus arrives between 5 minutes early and 9 minutes late.
Therefore, it defined the distribution as X~U[-5,9].However, the markscheme defines it as X~U[0,14]. As a result, I got everything wrong, except part d.Do you think I would've received the marks anyway? How would you proceed with this question?

Maybe the M1 marks, but x is considered from 5 minutes earlier than it's due time, so there's no need to go into negative values.
7. (Original post by JASISR)
How I remember skewness is by writing the values in the order MEAN MEDIA AND MODE
if mean<median<mode, it is negatively skewed because the pointy part of the inequality is pointing in the negative direction (think of a number line, the negative values are to the left of zero) and if MEAN>MEDIAN>MODE, then it is positively skewed as the pointy parts points in the positive direction. (Just my own odd way of remembering it
I like that
8. (Original post by MaxWalker1)
I like that
thanks MaxWalker1
9. (Original post by apzoe)
I just found an interesting question (January 2002).It says the bus arrives between 5 minutes early and 9 minutes late.
Therefore, it defined the distribution as X~U[-5,9].However, the markscheme defines it as X~U[0,14]. As a result, I got everything wrong, except part d.Do you think I would've received the marks anyway? How would you proceed with this question?

It specifies in the Q that X is the time after 7:55 and because the bus is expected to arrive at 8:00, then 5 mins before that would be 7:55. At 7:55 the time would be 0 mins after 7:55 hence why they defined it as 0

9 mins after the expected arrival time at 8:00 would be 8:09 which is 14 mins after 7:55
10. When they ask for 'the probability you incorrectly reject the null hypothesis' they are always asking for the actual significance level, right?
11. (Original post by SeanFM)
Let's see with an example.

If we had pdf = x for 0 <= x < 0.5 and 1-x for 0.5 <= x < 1 then the mode is 0.5.

The CDF is x^2/2 for 0 <= x < 0.5 and x - (x^/2) + (1/8) for 0.5 =< x < 1 so the median is indeed the mode.

The mean is integral of x between 1 and 0 which is 0.5.. so mode = median = mean. So we have shown that there is in fact no skew when all 3 of these things are true

But when there is skew, the mode is not equal to the median, which is not equal to the mean.

But you are right
Ahh thank you, this helps clear it up!

(Original post by RetroSpectro)
For anyone struggling to remember skews in a pdf use:

PMQU

P - M < Q < U

Positive - mode < Quartile 2 (median) < mu (mean)

and for a negative skew its the opposite direction
Smart way to learn it. But isn't is mean<median<mode instead of mode<median<mean?
12. continuity corrections people
13. I hope we don't get any definition questions
14. (Original post by RetroSpectro)
It specifies in the Q that X is the time after 7:55 and because the bus is expected to arrive at 8:00, then 5 mins before that would be 7:55. At 7:55 the time would be 0 mins after 7:55 hence why they defined it as 0

9 mins after the expected arrival time at 8:00 would be 8:09 which is 14 mins after 7:55
(Original post by NotNotBatman)
Maybe the M1 marks, but x is considered from 5 minutes earlier than it's due time, so there's no need to go into negative values.

Then would my way be correct if they didn't specify 7:55?

15. would i need to apply a contin correction
In the 3rd line of the mark scheme, why do they put it equal to 2? I understand the rest, any help is much appreciated Attachment 557073557075
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17. (Original post by apzoe)

Then would my way be correct if they didn't specify 7:55?
I think they would have to specify the variable

but if the specified variable was time after 8 am then you would be correct
18. (Original post by Yua)

would i need to apply a contin correction
No. You only apply the continuity correction when going from a discrete (binomial or poisson) random varriable to a continuous one (normal).
19. (Original post by undercxver)
Ahh thank you, this helps clear it up!

Smart way to learn it. But isn't is mean<median<mode instead of mode<median<mean?
Nope that would be a negative skew

20. (Original post by apzoe)
No. You only apply the continuity correction when going from a discrete (binomial or poisson) random varriable to a continuous one (normal).
thank you kind sir

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