Edexcel S2 - 27th June 2016 AM

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

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

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?

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

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?

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

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

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

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.

Ah my bad for rushing and not reading the question properly..

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

would i need to apply a contin correction

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?

No. You only apply the continuity correction when going from a discrete (binomial or poisson) random varriable to a continuous one (normal).