OCR A Level Further Mathematics MEI Statistics Minor Y432/01-16 Jun 2023 [Exam Chat]

Couldn’t do the last question and I cba to do the P(x=1) question. The rest I hope I did ok. Looking for at least a b maybe an a

what did everyone get for 1c , geometric question

0.8 somethinf

So I got 25/3 outside the bracket I think and then (n^2 - 38n something)

I got 25/6(n^2 -198n +9800)

This is correct

With the question before that, P[X<(n+100)/2]

I let n = 2k (will help generally and converted it later)

Range is n - 100 + 1 = n-99 (or 2k-99)
P[X<(k+50)] ~ i subbed in 2k
Because it's lower than (k + 50) it has to be k + 49 since it doesn't include that +1
P[X<]
So range of values it can take on is (k + 49) - (100) + 1 (1 extra because inclusive)
This leaves the range of possible values (or discrete "sides" is one way of thinking about it) to be:
k - 50
The probability of each "side" occurring is 1/(2k - 99). The formula for calculating the cumulative probability is [no. possible "sides" x 1/prob]
(k-50)/(2k-99).
Now converting back in terms of n:
0.5n-50/n-99 = n-100/2n-198

what did you write for the pmcc not being valid q out of interest

That was a bit tricky, as the data did, in fact, look roughly elliptical.
The variables were between sawing and cutting (I think)

Those two variables are not independent, which is a requirement of PMCC, so I talked about how if you're good at sawing, you are likely to be good at cutting. ie variables are dependent.

I also threw in other terms just in case, like possibly not being from a Bivariate normal distribution (even though it was a random sample, it was a random sample of only competitors, which itself gives bias when trying to find a correlation for a population.

I said it looks roughly elliptical only because of the outlier at the top right, would that work
and one final one on q3 three spins of a fair dice, did we have to do 10 squared or 10 times the variance as it said the sum and it said state one assumption that you are making or something along those lines

Yeah I talked about that one outlier as well for that question.

For the assumption for manipulations of variance and mean, we always talk about how each event occurs independent of each other.

and since we were saying sum of 10 independent values of X, it was variance x 10 square rooted for finding the sd, correct?

Nah it was just times by 10. Reason being is the Var(x) +- Var(y) = Var(x) + Var(y)

So it was 10Var(X) then square rooted