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Original post by kyan.cheung

Oh, oops, misread your thing.

No, to find Var(X^{2}) you find E[X^{4}]-(E[X^{2}])^{2} so the expression should be 176.9-8.9^{2}.

No, to find Var(X

Yh that’s what I got

Original post by drake12334

Dyu think 235-240 is a good estimate for grade boundaries FS1,FM1

Yes would say probably ~235 mark

Here is my solution to 7 c - I did this after the exam (after losing all 14 marks for the hard bits)

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\[ x \sim b(n, \frac{1}{5}) \]

\[ x \rightarrow \text{\# wins} \]

\[

\text{winnings} = \mathbb{e}\left( \sum_{i=1}^{x} 2i \right) = \mathbb{e} \left( 2 \sum_{i=1}^{x} i \right)

\]

\[

\text{winnings} = \mathbb{e} \left( x(x+1) \right)

\]

\[

= \mathbb{e} \left( x^2 + x \right)

\]

\[

= \mathbb{e} \left( x^2 \right) + \mathbb{e} (x)

\]

\[

\text{var}(x) = \mathbb{e} \left( x^2 \right) - (\mathbb{e} (x))^2

\]

\[

\mathbb{e} (x^2) = \text{var}(x) + (\mathbb{e} (x))^2

\]

\[

\text{winnings} = \mathbb{e} \left( x(x+1) \right) = np(1-p) + n^2p^2 + np

\]

\[

\text{winnings} = n \cdot \frac{1}{5} \cdot \frac{4}{5} + n^2 \cdot \frac{1}{25} + \frac{1}{5} n

\]

\[

= \frac{1}{25} n^2 + \frac{4n}{25} + \frac{5n}{25}

\]

\[

\text{profit} = \frac{1}{25} n^2 + \frac{9n}{25} - n = \frac{1}{25}n^2 + \frac{9n}{25} - \frac{25n}{25}

\]

\[

\text{profit} = \frac{1}{25} n^2 - \frac{16n}{25} = \frac{1}{25} (n^2 - 16n)

\]

\end{document}

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\[ x \sim b(n, \frac{1}{5}) \]

\[ x \rightarrow \text{\# wins} \]

\[

\text{winnings} = \mathbb{e}\left( \sum_{i=1}^{x} 2i \right) = \mathbb{e} \left( 2 \sum_{i=1}^{x} i \right)

\]

\[

\text{winnings} = \mathbb{e} \left( x(x+1) \right)

\]

\[

= \mathbb{e} \left( x^2 + x \right)

\]

\[

= \mathbb{e} \left( x^2 \right) + \mathbb{e} (x)

\]

\[

\text{var}(x) = \mathbb{e} \left( x^2 \right) - (\mathbb{e} (x))^2

\]

\[

\mathbb{e} (x^2) = \text{var}(x) + (\mathbb{e} (x))^2

\]

\[

\text{winnings} = \mathbb{e} \left( x(x+1) \right) = np(1-p) + n^2p^2 + np

\]

\[

\text{winnings} = n \cdot \frac{1}{5} \cdot \frac{4}{5} + n^2 \cdot \frac{1}{25} + \frac{1}{5} n

\]

\[

= \frac{1}{25} n^2 + \frac{4n}{25} + \frac{5n}{25}

\]

\[

\text{profit} = \frac{1}{25} n^2 + \frac{9n}{25} - n = \frac{1}{25}n^2 + \frac{9n}{25} - \frac{25n}{25}

\]

\[

\text{profit} = \frac{1}{25} n^2 - \frac{16n}{25} = \frac{1}{25} (n^2 - 16n)

\]

\end{document}

(edited 1 month ago)

Mm I reckon the mark allocation would be something like (of course, just a guess):

B1 for B(n, 1/5)

B1 for the arithmetic series sum

M1 for writing profit in terms of expectation

M1 for quoting Var(X) formula

A1 for valid E[W^2]

A1cso for getting the written value

Based off this, I reckon you've got the first B1 mark.

B1 for B(n, 1/5)

B1 for the arithmetic series sum

M1 for writing profit in terms of expectation

M1 for quoting Var(X) formula

A1 for valid E[W^2]

A1cso for getting the written value

Based off this, I reckon you've got the first B1 mark.

Original post by sssdssss

Yes would say probably ~235 mark

What about for an A, 200?

Original post by kyan.cheung

Mm I reckon the mark allocation would be something like (of course, just a guess):

B1 for B(n, 1/5)

B1 for the arithmetic series sum

M1 for writing profit in terms of expectation

M1 for quoting Var(X) formula

A1 for valid E[W^2]

A1cso for getting the written value

Based off this, I reckon you've got the first B1 mark.

B1 for B(n, 1/5)

B1 for the arithmetic series sum

M1 for writing profit in terms of expectation

M1 for quoting Var(X) formula

A1 for valid E[W^2]

A1cso for getting the written value

Based off this, I reckon you've got the first B1 mark.

Thank you!!!

Original post by nazarov

Did anyone do nth derivative of Gx(t) evaluated at 0, divided by n! to find probabilities?

Yeah I did this

Original post by nazarov

Nice, and do you guys think it will be 0 or 1 marks for getting 20 instead of 16 in 7c) using the wrong method

I think anyone who got 20 set up the correct distribution and summed the arithmetic series, so I'm guessing at least 2/6.

Original post by Talkative Toad

Edexcel A Level Further Mathematics Paper 3B (9FM0 3B) - 14th June 2024 [Exam Chat]

Welcome to the exam discussion thread for this exam. Introduce yourself! Let others know what you're aiming for in your exams, what you are struggling with in your revision or anything else.

Wishing you all the best of luck.

General Information

Date/Time: 14th June 2024/ PM

Length: 1hr 30 mins

Resources:

Edexcel A Level Further Mathematics

Welcome to the exam discussion thread for this exam. Introduce yourself! Let others know what you're aiming for in your exams, what you are struggling with in your revision or anything else.

Wishing you all the best of luck.

General Information

Date/Time: 14th June 2024/ PM

Length: 1hr 30 mins

Resources:

Edexcel A Level Further Mathematics

Anyone doing 3A and 3B (FS1 + FP1)? What do you think the grade boundaries might be based on the 3 exams so far?

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!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)=203/255

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 forgetting contingency tables and struggling on question 7, I think grade boundaries will be ~52 for A*.

1.

(6 marks total)

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)=203/255

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 forgetting contingency tables and struggling on question 7, I think grade boundaries will be ~52 for A*.

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

(edited 1 month ago)

i got 20% higher in mocks for fs1 over fm1 and now somehow in the actual exact i probably got 20%+ higher in fm1 than fs1

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!

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)=203/255

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 forgetting contingency tables and struggling on question 7, I think grade boundaries will be ~52 for A*.

1.

(6 marks total)

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)=203/255

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 forgetting contingency tables and struggling on question 7, I think grade boundaries will be ~52 for A*.

Original post by Stanleeee

I think 6b is 203/256? Because you have 27/4 * 1/64 for t^2 term so u can’t really have 255 on denominator

Your right he made a typo I presume

Original post by Supergremlin

Your right he made a typo I presume

Thanks for the unofficial mark scheme

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

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

Original post by th0mas006

What do you think the grade boundaries for furthers stats 1 and decision 1 will be?

wondering as well but i’d say 240 just to be extra safe

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