Edexcel IAL Statistics 2 Jan 25th 2017

Hello,

Did anyone here appeared for this exam today? We can discuss answers here since I don't see any other thread.

Funny story, I was the only registered candidate appearing for exam so I was alone with two invigilators, kinda weird but peaceful and quite.

Good luck, if you have any questions then let me know (I sat S2 ages ago... with only
One invigilator in the room 😂😂😂)

There was this question I am gonna write it out excuse me because it's blurry.

Piece of cloth has an average 2 defects over* 5m^2

A retailer buys 12 cloths of 15m^2
Find the probability that exactly half of these pieces of cloth has atleast 7 defects.

This was a 5 marks question I made some words up but the value and context of the question is exactly same. How would you go about doing this, my answers was either 0.7892 or 0.1709. I don't remember which was it.

I do remember a few more high marks question if you don't mind me asking let me know.

If average in 5m is 2 then in 15 it is 6.

It then becomes a binomial distribution question mixed with Poisson.

Probability of X, the event that there are at least 7 faults in a cloth, is found using Poisson. Let's call this probability p.

Then Y, the probability of exactly half of 12 cloths having 7 or more faults (in essence, 12 independent X's) can be modelled by a bionomial distribution with Y~B(12,p) and you are interested in P(Y=6).

If you have questions that can be explained just like that without going into calculations then you can ask.

Great that is exactly how I did it!

Yeah an other question

A farmer claims 96% of his seeds germinate.
In a random sample of 75 seeds, 66 seeds germinated.Use a suitable approximation to test at a 1% level of significance whether or not farmer is overstating the no. of his seeds germinate.

x~B(75, 0.96)
Y~N(72, 2.88)

H0p=0.96 ; H1p<0.96

P(X=<66) = P(Y<66.5) and took it from there. Does this working look correct to you? This was worth 7 marks I just want to be assured no silly mistakes were made.

Yes, looks like full marks to me providing conclusions were stated as they usually are in mark schemes

I let X=the number of seeds that didn't germinate and used X~B(75,0.04), then approximation X~Po(3).
I think it is not very suitable to use normal approximation because p is not close to 0.5.
BTW I rejected H0, is that true?

I let X=the number of seeds that didn't germinate and used X~B(75,0.04), then approximation X~Po(3).
I think it is not very suitable to use normal approximation because p is not close to 0.5.
BTW I rejected H0, is that true?

I let X=the number of seeds that didn't germinate and used X~B(75,0.04), then approximation X~Po(3).
I think it is not very suitable to use normal approximation because p is not close to 0.5.
BTW I rejected H0, is that true?