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Original post by Revi64

yeah i got this asw could someone confirm which is right?

it was greater than 5 not greater than or equal to im pretty sure. so P(X>=7) = 729/4096

Original post by MrMichi

what are we assuming is an A* in this paper all?

Probably around 55/75

Original post by seb9911

it was greater than 5 not greater than or equal to im pretty sure. so P(X>=7) = 729/4096

Yh but it would be 6-1 bcuz of what is said earlier no? Not 6+1

Original post by supergremlin

yh but it would be 6-1 bcuz of what is said earlier no? Not 6+1

p(x1+x2>5)

= p(y-1>5)

= p(y>6)

=p(y>=7)

Original post by Supergremlin

Q6d ) I got 81/256 for the probability of >=5

As Y = X1 + X2 +1

X1 +X2 = Y-1

Love to hear how you got 729/4096

As Y = X1 + X2 +1

X1 +X2 = Y-1

Love to hear how you got 729/4096

Original post by Supergremlin

Q6d ) I got 81/256 for the probability of >=5

As Y = X1 + X2 +1

X1 +X2 = Y-1

Love to hear how you got 729/4096

As Y = X1 + X2 +1

X1 +X2 = Y-1

Love to hear how you got 729/4096

yh it I got 729/4096 and it was >5 not >=5. so I basically did y-1 = x1 + x2 and then the pgf of y-1 is 1/(4t+9) (I forgot the denominator) and then I expanded using binomial expansion to find the coefficients up to t^5. and then I did P(x1+x2 > 5) = 1-p(x<=5)

Original post by yh21

what did everyone get for the cell you had to calculate was it the one with frequency of 3 btw

nah was bottom right as the expected frequency was the lowest

Original post by kettleboy

nah was bottom right as the expected frequency was the lowest

surely in context of the question he couldn’t have known that the bottom right has the lowest expected frequency?

Because he was selecting a cell in order to check the expected frequency only knowing the observed frequency, so surely 3 was right. that was my thinking anyway

Original post by Zacd25

surely in context of the question he couldn’t have known that the bottom right has the lowest expected frequency?

Because he was selecting a cell in order to check the expected frequency only knowing the observed frequency, so surely 3 was right. that was my thinking anyway

Because he was selecting a cell in order to check the expected frequency only knowing the observed frequency, so surely 3 was right. that was my thinking anyway

you choose the one with the lowest row total and the lowest column total which dont need maths to identify as expected = rowxcolumn / total

Original post by seb9911

p(x1+x2>5)

= p(y-1>5)

= p(y>6)

=p(y>=7)

= p(y-1>5)

= p(y>6)

=p(y>=7)

Your right thank you for clarifying

literally i was on track for a good A* but from today I'm gonna scrape it by 5-10 marks

Original post by drake12334

literally i was on track for a good A* but from today I'm gonna scrape it by 5-10 marks

Original post by sssdssss

What mark you think you got overall?

244 complete worse case

Original post by drake12334

244 complete worse case

Original post by Solysoly

I picked the one with 3 maybe there was a smaller one or another one with cell value 3?

I believe it was the bottom right cell, as that was the cell with the lowest probability. When working out Expected values.

Original post by kyan.cheung

Hi all, I've recorded the score distributions and my answers to each questions. Please tag me if you notice any issues!

1. (6 marks total)

a. (3 marks) Var(X)=4.9

b. (3 marks) Var(X^2)=97.69

2. (7 marks total)

ai. P(X=3) = 0.2090

aii. P(X<2)=0.3084 (2 marks total)

b. (4 marks) The Critical region is A<=1 or A>=12

c. (1 mark) P(Type I)=0.03744

3. (6 marks total)

a. (1 mark) Hypotheses - H0 colour and year group not associated, H1 colour and year group are associated

bi. Black Y10-12 because they are respectively the least frequent favourite colour and year group

ii. 5.383 (2 marks total)

c. (3 marks) (5-1)(3-1)=8 degrees of freedom with critical value 20.090, so reject H0 and conclude there is association between year group and favourite colour

4. (12 marks total)

a. (1 mark) X~Geo(1/6)

b. (2 marks) Show P(X<=3)=91/216

c. (5 marks) E[x]=6, Var(X)=30 -> Var(Xmean)30/64, P(5.6<Xmean<7.2)=0.6806

d. (4 marks) P(X>=16)=test value=0.064905...>0.05 so dont reject H0

5. (10 marks total)

a. (1 mark) H0 p=0.03 and H1 p>0.03

b. (2 marks) Size of test a 0.1821

c. (3 marks) Size of test b 0.2083

di. Power of a 0.7232

ii. We expect 94.877 compontents tested with test b (3 marks total)

e. Use test A as it has lower size and higher power (1 mark)

6. (16 marks total)

a. (6 marks) Variance 6

b. (4 marks) P(X<=2)=203/256

c. (4 marks) Y is a geometric distribution

d. (2 marks) P(X1+X2>5)=729/4096

7. (18 marks total)

a. (2 marks) Sam has prob 0.5886 of winning >=4 times

b. (4 marks) Don't remember what it asked but the answer was 0.04364 apparently

c. (6 marks) Show sam expects to win 1/25(n^2-16n). Sketch - basically set W~B(n, 1/5), then expected winnings is E[w^2+w-n] by sum rule. Then use E[w^2]=Var(W)+E^2[w]=4n/25+n^2/25 and simplify.

d. (4 marks) Sam has probability 0.3518 of not losing money

e. (4 marks) The other person has expected profit r^2-4r

RE predictions: with many people struggling on question 7, I think grade boundaries will be ~55 for A*.

1. (6 marks total)

a. (3 marks) Var(X)=4.9

b. (3 marks) Var(X^2)=97.69

2. (7 marks total)

ai. P(X=3) = 0.2090

aii. P(X<2)=0.3084 (2 marks total)

b. (4 marks) The Critical region is A<=1 or A>=12

c. (1 mark) P(Type I)=0.03744

3. (6 marks total)

a. (1 mark) Hypotheses - H0 colour and year group not associated, H1 colour and year group are associated

bi. Black Y10-12 because they are respectively the least frequent favourite colour and year group

ii. 5.383 (2 marks total)

c. (3 marks) (5-1)(3-1)=8 degrees of freedom with critical value 20.090, so reject H0 and conclude there is association between year group and favourite colour

4. (12 marks total)

a. (1 mark) X~Geo(1/6)

b. (2 marks) Show P(X<=3)=91/216

c. (5 marks) E[x]=6, Var(X)=30 -> Var(Xmean)30/64, P(5.6<Xmean<7.2)=0.6806

d. (4 marks) P(X>=16)=test value=0.064905...>0.05 so dont reject H0

5. (10 marks total)

a. (1 mark) H0 p=0.03 and H1 p>0.03

b. (2 marks) Size of test a 0.1821

c. (3 marks) Size of test b 0.2083

di. Power of a 0.7232

ii. We expect 94.877 compontents tested with test b (3 marks total)

e. Use test A as it has lower size and higher power (1 mark)

6. (16 marks total)

a. (6 marks) Variance 6

b. (4 marks) P(X<=2)=203/256

c. (4 marks) Y is a geometric distribution

d. (2 marks) P(X1+X2>5)=729/4096

7. (18 marks total)

a. (2 marks) Sam has prob 0.5886 of winning >=4 times

b. (4 marks) Don't remember what it asked but the answer was 0.04364 apparently

c. (6 marks) Show sam expects to win 1/25(n^2-16n). Sketch - basically set W~B(n, 1/5), then expected winnings is E[w^2+w-n] by sum rule. Then use E[w^2]=Var(W)+E^2[w]=4n/25+n^2/25 and simplify.

d. (4 marks) Sam has probability 0.3518 of not losing money

e. (4 marks) The other person has expected profit r^2-4r

RE predictions: with many people struggling on question 7, I think grade boundaries will be ~55 for A*.

bro why you’re even doing a levels ur in uni aren’t u😭😭

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