aqa further stats

[limber ore]

How do I do part a?

[frank snow]

Have you considered ig modelling it like a poissoin distribuition and using your knowledge of that figuring out the corresponding Variance and Mean

[limber ore]

this is the mark scheme

i dont really understand it

[limber ore]

i kind of get it now

@frank snow can you help me with this

how do you got from 3rd line to 4th

[frank snow]

Im out rn but ill respond when i get home for sure

[limber ore]

Okk

[frank snow]

Ahh ok so it modelled it as a uniform distribuition and then using the formula to calculate variance and expectation

[frank snow]

This comes from the fact that no events happening in t would be e^-6t. Are you confused as to why this is the case?

[limber ore]

ye

why does no events happening imply that its e^-6t

like i understand if you had the probablility density function 6e^-6x

and you integrated it to get the cumulative probability function youd get 1-e^-6t

but they didnt do that

what relevance does the P(no events in t) have

[frank snow]

1-P(no events in t) = P(T<t)

[limber ore]

ye

[frank snow]

Im not too sure tbh i thought the distribuition is Poissoin but im either wrong or cant remember how to manipulate stuff

[limber ore]

dw

i asked a few people and they said they werent sure

i doubt it matters to much

[frank snow]

It could be something to do with the integral between certain bound being equal to 1

[limber ore]

obtw i figued it out

P(no events) = P(X=0)

since its defined as a possion distribution at the start

x=0 in that

e^-6t

oh shit

the ms doesnt really go into the possion bit

but its from here

[frank snow]

Ohh fair i thought it was poissoin related just havent done enough revision off this topic in while. Nice Q tho made me realise i need to revise