OCR MEI Statistics 1 (S1)

Im really quite happy with most of these. There seems to be some differing answers on the E(x) and V(x) though. I believe E(x) was 2.6. Then E(x^2)=8, so V(x)=(8-2.6^2).

Also on the midrange question, I put that it could very well be the answer as the midrange would have to be between 1750 and 3000.

I have to say, I drew A WHOLE LOT of probability trees for this paper, a lot more of those questions then expected... I know you were meant to use binomials and stuff, so I put these in answer I knew the answers from the probability tree :wink:

Pretty sure that's what I got for Var and E, 2.6 and 8 certainly ring a bell.

Reply 221

Axion

15

E(x) was 2.6 for sure.

then Var(x) = all the numbers minus E(x)^2

which was 8 - 2.6^2

therefore Var(x) = 1.24

Reply 222

Polioz

14

all the As nooblets in my college found this paper extremely difficult, even some a2 people. Im hoping for low grade boundaries

I dont understand question 1 iii though. it said what is the probability of having atleast 1 of each colour , so the choice is either RRB ,RBR, BBR,BRB, but I got a different answer?

Reply 223

Ldbright

Original post by Polioz

all the As nooblets in my college found this paper extremely difficult, even some a2 people. Im hoping for low grade boundaries

I dont understand question 1 iii though. it said what is the probability of having atleast 1 of each colour , so the choice is either RRB ,RBR, BBR,BRB, but I got a different answer?

There could also be BRR, or RBB. However, what I did was just do (1-BBB-RRR) as all others would have 1 of each at least. So in the End what I got was 1-(30/50*29/49*28/48)-(20/50*19/49*18/48)

Reply 224

nicknickmac

For the question about putting callers on hold... What did people get for all 3 parts.
A) Prob that exactly 29 are put on hold
B) Prob that at least 29 are put on hold.
C) 10 samples, in how many are at least 29 put on hold.

x

Reply 225

Ldbright

Original post by nicknickmac

For the question about putting callers on hold... What did people get for all 3 parts.
A) Prob that exactly 29 are put on hold
B) Prob that at least 29 are put on hold.
C) 10 samples, in how many are at least 29 put on hold.

x

A) 30C29*0.85^29*0.15= 0.04040

B) 1-(30C29*0.85^29*0.15)-0.85*30= 0.04803

C) 10*0.04803=0.4803

Anyone who sees this, all the answers are on page 10 by "deceptively"

Reply 226

Polioz

14

damn I thought I came out with 68/72, looking over markscheme I think I lost 11 marks. What are peoples thoughts on grade boundaries? I think this was harder than last years one which was 53/72 for A.

hoping 61/72 will be 90+ :s-smilie:

Reply 227

louci

7

Original post by Polioz

damn I thought I came out with 68/72, looking over markscheme I think I lost 11 marks. What are peoples thoughts on grade boundaries? I think this was harder than last years one which was 53/72 for A.

hoping 61/72 will be 90+ :s-smilie:

i know i am late but can u tell me where the markscheme is ?

Reply 228

EsmeA

Original post by KageKitsune

I did that too.. Any teachers on here to give the correct answer please? ^^

I did the 18 combinations too, I don't understand why it's wrong, can anyone explain?

The way I imagined it was that even though it was three ones, (1,1,1), they still could have arrived in any order, which was 3P3 if you imagined it to be "1.1, 1.2, 1.3" or "1.3,1.2,1.1" if you see what I mean. I thought that as each group of three there'd be six combinations, making a total of 18 instead of ten.