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# Official Thread: OCR MEI S2/M1 Watch

1. (Original post by groovy_q)
Did anyone reject h0 for the rank test and the 11 marker, and accept h0 for the chi squared test?
I'm terrible at remembering what happened in exams, but I do seem to remember it going reject; accept; reject. So, assuming that's not false memory, I think so!
But again, I just remember there being 2 of 1 and 1 of another so this is vague af.

Edit: Unofficial MS says accept; accept; reject, and I got the same numbers as him so I think that might be the way acc. Yey memories.
2. Does anyone have the m1 june 15 paper for revision purposes?
3. (Original post by Leechayy)

Posted from TSR Mobile
Thanks!

If this post gets a few reps I'll do a full unofficial mark scheme (it's worth doing it if helps more than just me )
4. If you guys want a full in depth unofficial mark scheme for S2 let me know (or rep the post above to let me know indirectly ^^)
5. http://www.mei.org.uk/files/2000papers/m105ja.pdf

can someon explain 2iii to me, how are the angles 120 30 and 30
6. (Original post by ComputerMaths97)
Thanks!

If this post gets a few reps I'll do a full unofficial mark scheme (it's worth doing it if helps more than just me )

Posted from TSR Mobile
7. (Original post by ComputerMaths97)
What do you mean, like the difference between 0.4567 and 0.4568 for example?

I reckon boundaries will be higher. Was very similar to last year, a little easier in some aspects. The hardest part (i.e the part that I reckon most people lost marks on) was the last part, which was in last years paper and therefore more people would've been prepared for it.
Yep thats what I mean, would I lose any marks do you really reckon it will be 65+, I thought there was a lot for silly errors and not sure about the monotonic question
8. (Original post by 11234)
Yep thats what I mean, would I lose any marks do you really reckon it will be 65+, I thought there was a lot for silly errors and not sure about the monotonic question
You will lose marks if you made a mistake usually yes. I reckon so, was a very standard paper. Yeah that monotonic (or no assumption?) question will just mean that the 100 UMS boundary will probs be 71
9. Here is my unofficial S2 MEI OCR June 2016 Mark Scheme

Spoiler:
Show

- This is me comparing the questions to previous years, and using their mark scheme. So if you disagree, I may have made a calculation error. Just let me know if you think anything should be disputed, but all methods are from previous exam boards.
- This is unofficial (yeah some people will think this is the real thing)
- If you're cheating for your mock (to you lot seeing this in 2017), remember that you gain nothing from cheating, and this is not official. S2 is mainly about learning past paper material, so you're only making it harder for yourself by not revising properly for your mock. But who needs a lecture from me ay
- But other than that, enjoy (Be sure to rep if you agree/ appreciate the couple hours I spend making this )
S2 June 2016 Unofficial Mark Scheme:
Spoiler:
Show

Q1)
i) Scatter diagram (3 marks)
Linear scales and axis labelled "Quantity of fertiliser kg/hectare" and "Percentage of population living in rural areas (%)" or similarly suitable = (1 mark)
All 11 points plotted correctly = (2 marks)
(Only 9 (or more) points plotted correctly then 1 mark instead of the 2 from above)
Note: Does not matter which axis is which variable

ii) Why pmcc test might not be valid (2 marks)
The scatter diagram does not appear to be roughly elliptical = (1 mark)
So the population may not have a bivariate normal distribution = (1 mark)

iii) Find spearmans rank correlation coefficient (5 marks)
x rank: 4 , 1 , 6 , 5 , 11 , 8 , 7 , 2 , 10 , 9 , 3
y rank: 8 , 6 , 3 , 7 , 1 , 5 , 4 , 11 , 2 , 9 , 10 (these two rows correct = 1 mark)
d^2....: 16,25,9,4,100,9,9,81,64,0,49 (row correct = 1 mark)

sum of d^2 = 366 (correct answer only = 1 mark)

r = 1 - [6 x sigma(d^2)]/n(n^2 -1)
= 1 - 6(366)/11(120)
= 1- 2196/1320 (method = 1 mark)
= -0.6636 (answer = 1 mark - allowed to 2.d.p or better)

iv) Do (negative association) test at 5% level (6 marks)
H0: No association between the quantity of fertiliser used and the percentage of the population who live in rural areas in the population of countries
H1: Negative association ...

(1 mark for both hypothesis)
(1 mark for both in context)
(1 mark for "in the population of countries"

Critical value = 0.7091 (1 mark)
Since |-0.6636| < 0.7091, result not significant, therefore accept H0 (1 mark)
Conclusion in context + non assertive i.e There is insufficient evidence to suggest that there is some association between....

v) Explain the term "1% significance level" (1 mark)
The probability of rejecting the null hypothesis when it is in fact true is 1%/0.01 (1 mark)

vi) State modelling distribution assumptions for spearmans test to be valid (1 mark)
*This has never come up in an exam before. After hours of research, it's clear it's one of two answers, I'm sorry I'm unable to give a certain answer*
(Answer option 1) There are none, since spearmans rcc is calculated using rank not actual values and is therefore not concerned by actual distribution of values, just their rank
(Answer option 2) For the spearmans test to work, the underlying relationship must be monotonic.

The only reason it's not certainly number 2) is because this phrase only exists in one mei document that's only available to teachers (not in textbook, on integral, nor in past papers) and I have quoted it. Notice is says "relationship", which raises concerns to me. My teacher had never heard of monotonic so I can't be too sure.

2)
i) Define 'independently' and 'randomly' in context (2 marks)
Independently means that the occurence of one mutation does not affect the probability of another mutation occuring (1 mark)
Randomly means that the mutations are unpredictable (1 mark)

ii) Find P(X=1) (2 marks)
Binomial (20,0.012)
Therefore P(X=1) = 20C1 x 0.012 x (0.988)^19 (1 mark)
= 0.1908 (1 mark)

iii) Why is Poisson suitable as approximation (2 marks)
Since it is binomially distributed:
n is large (1 mark)
p is small (1 mark)

iv)
A) Find P(X=2) (3 marks)
lambda = np = 500 x 0.012 = 6 (1 mark)
P(X=2) = [e^-6 x 6^2]/2 (1 mark)
= 0.0446 (1 mark)

B) Find P(X>=2) (2 marks)
P(X>=2) = 1-P(X<=1) (1 mark)
=1 - 0.0174
= 0.9826 (1 mark)

v) Use suitable approximating distribution to find P(X >= 650) (5 marks)
mean = np = 50000x0.012 = 600 (1 mark)
variance = npq = 50000x0.012x0.988 = 592.8 (1 mark)
P(X >= 650) = P(Z > (649.5-600)/sqrt(592.8)) (1 mark for continuity correction)
= P(Z > 2.033) = 1 - phi(2.033) (1 mark for this CORRECT ONLY)
= 1 - 0.9789
= 0.0211 (1 mark)

3)
i) Find P(50000 < X < 55000) (3 marks)
= P([50000-50600]/3400 < Z < [55000 - 50600]/3400)
= P( -0.1765 < Z < 1.2941)
= phi(1.2941) - (1-phi(0.1765)) (1 mark for correct form here)
= 0.9022 - (1 - 0.5701) (1 mark for either of these values read of tables correctly)
= 0.4723 (1 mark)

ii) Claims that at least 95% last over 45000 hours, is he correct? (3 marks)
P(X>45000) = P(Z > [45000-50600]/3400)
= P(Z > -1.6471)
= phi(1.6471) (1 mark for this correct)
= 0.9502 (1 mark)
Therefore since 0.9502 > 0.95, claim is valid (1 mark for specifying what you just calculated means)

iii) Find h such that 99.9% of them last h or more hours (3 marks)
P(X >= h) = 0.999
......
(h-50600)/3400 = - phi(0.999)
(h-50600)/3400 = -3.090 (1 mark for +/- 3.090, 1 mark for equation in this form even if incorrect)
h = 40094 hours (1 mark)

iv) P(Y < 60000) = 0.6 and P(Y > 50000) = 0.9, find mean (x) and standard deviation (s) of Y ( 5 marks)
P(Y < 60000) = 0.6
P(Z < (60000-x)/s) = 0.6
(60000 - x) / s = inverse-phi (0.6)
60000 - x = 0.2533s (1 mark for equation in this form even if incorrect)
P(Y > 50000) = 0.9
......
50000 - x = -1.282s (1 mark for equation in this form even if incorrect)
(1 mark for either inverse-phi(0.6) = 0.2533 or inverse-phi(0.9) = 1.282 correct)
s = 6513 (or 6513.4 or 6513.39) (1 mark)
x = 58350 (1 mark)

v) Draw normal curves of X and Y on same axis (4 marks)
Two normal shapes, asymptotic behaviour and horizontal axis/values correctly labelled (1 mark)
Both means explicitely stated (1 mark)
Lower max height for Y (1 mark)
Visibly greater width for Y (1 mark)
Note: Y should be to the right of X and if either of the curves are evidently not symmetrical then a mark is lost)

4)
i) Find expected frequency of female agree, and verify test statistic of 0.6831 (3 marks)
Expected frequency = (29x42)/80 = 15.225 (1 mark)
Contribution = (12-15.225)^2 / 15.225 (1 mark for attempt at (observed-expected)^2/expected)
= 0.6831 (1 mark)

ii) Do 5% sig level test of association (6 marks)
H0: No association between the sex of the student and their attitude towards mathematics
H1: Some association between .... (1 mark for both correct)
Sum of chi-squared = 5.3236 (1 mark)
2 degrees of freedom (1 mark)
Critical value at 5% level = 5.991 (1 mark)
Since 5.3236 < 5.991, results not significant, accept H0 (1 mark)
Conclusion in context, non assertive (1 mark)

b) Do test at 5% level for increased mean (11 marks)
mean = sigma(x)/n = 373/60 = 6.217 (1 mark)
s = sqrt((sigma(x^2)-(sigma(x)^2)/n)/n-1)
= sqrt (2498-(373^2/60)/59)
= sqrt(3.037) (1 mark for method of finding sample s)
=1.7427 (1 mark for explicitely giving the value of s or s^2)

H0: mean = 5.64
H1: mean > 5.64 (1 mark for both correct)
When mean = the mean level of radioactivity in limpets for the population (1 mark for defining this)

Test statistic = (6.217-5.64)/(1.7427/sqrt(60) (1 mark for dividing s by sqrt(60))
= 2.5647 (1 mark for correct value)
5% one tail critical value = 1.645 (1 mark)

Since 2.5647 > 1.645 result is significant (1 mark)

Reject H0, accept H1 (1 mark)
Conclusion, non assertive, in context (1 mark)
Phew that took SO LONG Please let me know of any errors I made Would appreciate the reps

Let me know how you think you did! I think I got anywhere from 69/72 to 72/72 based on memory of my answers not being perfect
10. (Original post by ComputerMaths97)
Here is my unofficial S2 MEI OCR June 2016 Mark Scheme

Spoiler:
Show

- This is me comparing the questions to previous years, and using their mark scheme. So if you disagree, I may have made a calculation error. Just let me know if you think anything should be disputed, but all methods are from previous exam boards.
- This is unofficial (yeah some people will think this is the real thing)
- If you're cheating for your mock (to you lot seeing this in 2017), remember that you gain nothing from cheating, and this is not official. S2 is mainly about learning past paper material, so you're only making it harder for yourself by not revising properly for your mock. But who needs a lecture from me ay
- But other than that, enjoy (Be sure to rep if you agree/ appreciate the couple hours I spend making this )
S2 June 2016 Unofficial Mark Scheme:
Spoiler:
Show

Q1)
i) Scatter diagram (3 marks)
Linear scales and axis labelled "Quantity of fertiliser kg/hectare" and "Percentage of population living in rural areas (%)" or similarly suitable = (1 mark)
All 11 points plotted correctly = (2 marks)
(Only 9 (or more) points plotted correctly then 1 mark instead of the 2 from above)
Note: Does not matter which axis is which variable

ii) Why pmcc test might not be valid (2 marks)
The scatter diagram does not appear to be roughly elliptical = (1 mark)
So the population may not have a bivariate normal distribution = (1 mark)

iii) Find spearmans rank correlation coefficient (5 marks)
x rank: 4 , 1 , 6 , 5 , 11 , 8 , 7 , 2 , 10 , 9 , 3
y rank: 8 , 6 , 3 , 7 , 1 , 5 , 4 , 11 , 2 , 9 , 10 (these two rows correct = 1 mark)
d^2....: 16,25,9,4,100,9,9,81,64,0,49 (row correct = 1 mark)

sum of d^2 = 366 (correct answer only = 1 mark)

r = 1 - [6 x sigma(d^2)]/n(n^2 -1)
= 1 - 6(366)/11(120)
= 1- 2196/1320 (method = 1 mark)
= -0.6636 (answer = 1 mark - allowed to 2.d.p or better)

iv) Do (negative association) test at 5% level (6 marks)
H0: No association between the quantity of fertiliser used and the percentage of the population who live in rural areas in the population of countries
H1: Some association ...

(1 mark for both hypothesis)
(1 mark for both in context)
(1 mark for "in the population of countries"

Critical value = 0.7091 (1 mark)
Since |-0.6636| < 0.7091, result not significant, therefore accept H0 (1 mark)
Conclusion in context + non assertive i.e There is insufficient evidence to suggest that there is some association between....

v) Explain the term "1% significance level" (1 mark)
The probability of rejecting the null hypothesis when it is in fact true is 1%/0.01 (1 mark)

vi) State modelling distribution assumptions for spearmans test to be valid (1 mark)
*This has never come up in an exam before. After hours of research, it's clear it's one of two answers, I'm sorry I'm unable to give a certain answer*
(Answer option 1) There are none, since spearmans rcc is calculated using rank not actual values and is therefore not concerned by actual distribution of values, just their rank
(Answer option 2) For the spearmans test to work, the underlying relationship must be monotonic.

The only reason it's not certainly number 2) is because this phrase only exists in one mei document that's only available to teachers (not in textbook, on integral, nor in past papers) and I have quoted it. Notice is says "relationship", which raises concerns to me. My teacher had never heard of monotonic so I can't be too sure.

2)
i) Define 'independently' and 'randomly' in context (2 marks)
Independently means that the occurence of one mutation does not affect the probability of another mutation occuring (1 mark)
Randomly means that the mutations are unpredictable (1 mark)

ii) Find P(X=1) (2 marks)
Binomial (20,0.012)
Therefore P(X=1) = 20C1 x 0.012 x (0.988)^19 (1 mark)
= 0.1908 (1 mark)

iii) Why is Poisson suitable as approximation (2 marks)
Since it is binomially distributed:
n is large (1 mark)
p is small (1 mark)

iv)
A) Find P(X=2) (3 marks)
lambda = np = 500 x 0.012 = 6 (1 mark)
P(X=2) = [e^-6 x 6^2]/2 (1 mark)
= 0.0446 (1 mark)

B) Find P(X>=2) (2 marks)
P(X>=2) = 1-P(X<=1) (1 mark)
=1 - 0.0174
= 0.9826 (1 mark)

v) Use suitable approximating distribution to find P(X >= 650) (5 marks)
mean = np = 50000x0.012 = 600 (1 mark)
variance = npq = 50000x0.012x0.988 = 592.8 (1 mark)
P(X >= 650) = P(Z > (649.5-600)/sqrt(592.8)) (1 mark for continuity correction)
= P(Z > 2.033) = 1 - phi(2.033) (1 mark for this CORRECT ONLY)
= 1 - 0.9789
= 0.0211 (1 mark)

3)
i) Find P(50000 < X < 55000) (3 marks)
= P([50000-50600]/3400 < Z < [55000 - 50600]/3400)
= P( -0.1765 < Z < 1.2941)
= phi(1.2941) - (1-phi(0.1765)) (1 mark for correct form here)
= 0.9022 - (1 - 0.5701) (1 mark for either of these values read of tables correctly)
= 0.4723 (1 mark)

ii) Claims that at least 95% last over 45000 hours, is he correct? (3 marks)
P(X>45000) = P(Z > [45000-50600]/3400)
= P(Z > -1.6471)
= phi(1.6471) (1 mark for this correct)
= 0.9502 (1 mark)
Therefore since 0.9502 > 0.95, claim is valid (1 mark for specifying what you just calculated means)

iii) Find h such that 99.9% of them last h or more hours (3 marks)
P(X >= h) = 0.999
......
(h-50600)/3400 = - phi(0.999)
(h-50600)/3400 = -3.090 (1 mark for +/- 3.090, 1 mark for equation in this form even if incorrect)
h = 40094 hours (1 mark)

iv) P(Y < 60000) = 0.6 and P(Y > 50000) = 0.9, find mean (x) and standard deviation (s) of Y ( 5 marks)
P(Y < 60000) = 0.6
P(Z < (60000-x)/s) = 0.6
(60000 - x) / s = inverse-phi (0.6)
60000 - x = 0.2533s (1 mark for equation in this form even if incorrect)
P(Y > 50000) = 0.9
......
50000 - x = -1.282s (1 mark for equation in this form even if incorrect)
(1 mark for either inverse-phi(0.6) = 0.2533 or inverse-phi(0.9) = 1.282 correct)
s = 6513 (or 6513.4 or 6513.39) (1 mark)
x = 58350 (1 mark)

v) Draw normal curves of X and Y on same axis (4 marks)
Two normal shapes, asymptotic behaviour and horizontal axis/values correctly labelled (1 mark)
Both means explicitely stated (1 mark)
Lower max height for Y (1 mark)
Visibly greater width for Y (1 mark)
Note: Y should be to the right of X and if either of the curves are evidently not symmetrical then a mark is lost)

4)
i) Find expected frequency of female agree, and verify test statistic of 0.6831 (3 marks)
Expected frequency = (29x42)/80 = 15.225 (1 mark)
Contribution = (12-15.225)^2 / 15.225 (1 mark for attempt at (observed-expected)^2/expected)
= 0.6831 (1 mark)

ii) Do 5% sig level test of association (6 marks)
H0: No association between the sex of the student and their attitude towards mathematics
H1: Some association between .... (1 mark for both correct)
Sum of chi-squared = 5.3236 (1 mark)
2 degrees of freedom (1 mark)
Critical value at 5% level = 5.991 (1 mark)
Since 5.3236 < 5.991, results not significant, accept H0 (1 mark)
Conclusion in context, non assertive (1 mark)

b) Do test at 5% level for increased mean (11 marks)
mean = sigma(x)/n = 373/60 = 6.217 (1 mark)
s = sqrt((sigma(x^2)-(sigma(x)^2)/n)/n-1)
= sqrt (2498-(373^2/60)/59)
= sqrt(3.037) (1 mark for method of finding sample s)
=1.7427 (1 mark for explicitely giving the value of s or s^2)

H0: mean = 5.64
H1: mean > 5.64 (1 mark for both correct)
When mean = the mean level of radioactivity in limpets for the population (1 mark for defining this)

Test statistic = (6.217-5.64)/(1.7427/sqrt(60) (1 mark for dividing s by sqrt(60))
= 2.5647 (1 mark for correct value)
5% one tail critical value = 1.645 (1 mark)

Since 2.5647 > 1.645 result is significant (1 mark)

Reject H0, accept H1 (1 mark)
Conclusion, non assertive, in context (1 mark)
Phew that took SO LONG Please let me know of any errors I made Would appreciate the reps

Let me know how you think you did! I think I got anywhere from 69/72 to 72/72 based on memory of my answers not being perfect
Thankyou for writing all this out!

Assuming everything is correct here, I have 69-70 marks

For the 0.9502 one, I did it like this:

Phi^-1(0.05)= -1.645
-1.645 x 3400 + 50600 < 45000
But stupid me said it was invalid rather than valid so I lose my marks there

Other than that, I'm good

Posted from TSR Mobile
11. (Original post by ComputerMaths97)
Here is my unofficial S2 MEI OCR June 2016 Mark Scheme

Spoiler:
Show

- This is me comparing the questions to previous years, and using their mark scheme. So if you disagree, I may have made a calculation error. Just let me know if you think anything should be disputed, but all methods are from previous exam boards.
- This is unofficial (yeah some people will think this is the real thing)
- If you're cheating for your mock (to you lot seeing this in 2017), remember that you gain nothing from cheating, and this is not official. S2 is mainly about learning past paper material, so you're only making it harder for yourself by not revising properly for your mock. But who needs a lecture from me ay
- But other than that, enjoy (Be sure to rep if you agree/ appreciate the couple hours I spend making this )
S2 June 2016 Unofficial Mark Scheme:
Spoiler:
Show

Q1)
i) Scatter diagram (3 marks)
Linear scales and axis labelled "Quantity of fertiliser kg/hectare" and "Percentage of population living in rural areas (%)" or similarly suitable = (1 mark)
All 11 points plotted correctly = (2 marks)
(Only 9 (or more) points plotted correctly then 1 mark instead of the 2 from above)
Note: Does not matter which axis is which variable

ii) Why pmcc test might not be valid (2 marks)
The scatter diagram does not appear to be roughly elliptical = (1 mark)
So the population may not have a bivariate normal distribution = (1 mark)

iii) Find spearmans rank correlation coefficient (5 marks)
x rank: 4 , 1 , 6 , 5 , 11 , 8 , 7 , 2 , 10 , 9 , 3
y rank: 8 , 6 , 3 , 7 , 1 , 5 , 4 , 11 , 2 , 9 , 10 (these two rows correct = 1 mark)
d^2....: 16,25,9,4,100,9,9,81,64,0,49 (row correct = 1 mark)

sum of d^2 = 366 (correct answer only = 1 mark)

r = 1 - [6 x sigma(d^2)]/n(n^2 -1)
= 1 - 6(366)/11(120)
= 1- 2196/1320 (method = 1 mark)
= -0.6636 (answer = 1 mark - allowed to 2.d.p or better)

iv) Do (negative association) test at 5% level (6 marks)
H0: No association between the quantity of fertiliser used and the percentage of the population who live in rural areas in the population of countries
H1: Some association ...

(1 mark for both hypothesis)
(1 mark for both in context)
(1 mark for "in the population of countries"

Critical value = 0.7091 (1 mark)
Since |-0.6636| < 0.7091, result not significant, therefore accept H0 (1 mark)
Conclusion in context + non assertive i.e There is insufficient evidence to suggest that there is some association between....

v) Explain the term "1% significance level" (1 mark)
The probability of rejecting the null hypothesis when it is in fact true is 1%/0.01 (1 mark)

vi) State modelling distribution assumptions for spearmans test to be valid (1 mark)
*This has never come up in an exam before. After hours of research, it's clear it's one of two answers, I'm sorry I'm unable to give a certain answer*
(Answer option 1) There are none, since spearmans rcc is calculated using rank not actual values and is therefore not concerned by actual distribution of values, just their rank
(Answer option 2) For the spearmans test to work, the underlying relationship must be monotonic.

The only reason it's not certainly number 2) is because this phrase only exists in one mei document that's only available to teachers (not in textbook, on integral, nor in past papers) and I have quoted it. Notice is says "relationship", which raises concerns to me. My teacher had never heard of monotonic so I can't be too sure.

2)
i) Define 'independently' and 'randomly' in context (2 marks)
Independently means that the occurence of one mutation does not affect the probability of another mutation occuring (1 mark)
Randomly means that the mutations are unpredictable (1 mark)

ii) Find P(X=1) (2 marks)
Binomial (20,0.012)
Therefore P(X=1) = 20C1 x 0.012 x (0.988)^19 (1 mark)
= 0.1908 (1 mark)

iii) Why is Poisson suitable as approximation (2 marks)
Since it is binomially distributed:
n is large (1 mark)
p is small (1 mark)

iv)
A) Find P(X=2) (3 marks)
lambda = np = 500 x 0.012 = 6 (1 mark)
P(X=2) = [e^-6 x 6^2]/2 (1 mark)
= 0.0446 (1 mark)

B) Find P(X>=2) (2 marks)
P(X>=2) = 1-P(X<=1) (1 mark)
=1 - 0.0174
= 0.9826 (1 mark)

v) Use suitable approximating distribution to find P(X >= 650) (5 marks)
mean = np = 50000x0.012 = 600 (1 mark)
variance = npq = 50000x0.012x0.988 = 592.8 (1 mark)
P(X >= 650) = P(Z > (649.5-600)/sqrt(592.8)) (1 mark for continuity correction)
= P(Z > 2.033) = 1 - phi(2.033) (1 mark for this CORRECT ONLY)
= 1 - 0.9789
= 0.0211 (1 mark)

3)
i) Find P(50000 < X < 55000) (3 marks)
= P([50000-50600]/3400 < Z < [55000 - 50600]/3400)
= P( -0.1765 < Z < 1.2941)
= phi(1.2941) - (1-phi(0.1765)) (1 mark for correct form here)
= 0.9022 - (1 - 0.5701) (1 mark for either of these values read of tables correctly)
= 0.4723 (1 mark)

ii) Claims that at least 95% last over 45000 hours, is he correct? (3 marks)
P(X>45000) = P(Z > [45000-50600]/3400)
= P(Z > -1.6471)
= phi(1.6471) (1 mark for this correct)
= 0.9502 (1 mark)
Therefore since 0.9502 > 0.95, claim is valid (1 mark for specifying what you just calculated means)

iii) Find h such that 99.9% of them last h or more hours (3 marks)
P(X >= h) = 0.999
......
(h-50600)/3400 = - phi(0.999)
(h-50600)/3400 = -3.090 (1 mark for +/- 3.090, 1 mark for equation in this form even if incorrect)
h = 40094 hours (1 mark)

iv) P(Y < 60000) = 0.6 and P(Y > 50000) = 0.9, find mean (x) and standard deviation (s) of Y ( 5 marks)
P(Y < 60000) = 0.6
P(Z < (60000-x)/s) = 0.6
(60000 - x) / s = inverse-phi (0.6)
60000 - x = 0.2533s (1 mark for equation in this form even if incorrect)
P(Y > 50000) = 0.9
......
50000 - x = -1.282s (1 mark for equation in this form even if incorrect)
(1 mark for either inverse-phi(0.6) = 0.2533 or inverse-phi(0.9) = 1.282 correct)
s = 6513 (or 6513.4 or 6513.39) (1 mark)
x = 58350 (1 mark)

v) Draw normal curves of X and Y on same axis (4 marks)
Two normal shapes, asymptotic behaviour and horizontal axis/values correctly labelled (1 mark)
Both means explicitely stated (1 mark)
Lower max height for Y (1 mark)
Visibly greater width for Y (1 mark)
Note: Y should be to the right of X and if either of the curves are evidently not symmetrical then a mark is lost)

4)
i) Find expected frequency of female agree, and verify test statistic of 0.6831 (3 marks)
Expected frequency = (29x42)/80 = 15.225 (1 mark)
Contribution = (12-15.225)^2 / 15.225 (1 mark for attempt at (observed-expected)^2/expected)
= 0.6831 (1 mark)

ii) Do 5% sig level test of association (6 marks)
H0: No association between the sex of the student and their attitude towards mathematics
H1: Some association between .... (1 mark for both correct)
Sum of chi-squared = 5.3236 (1 mark)
2 degrees of freedom (1 mark)
Critical value at 5% level = 5.991 (1 mark)
Since 5.3236 < 5.991, results not significant, accept H0 (1 mark)
Conclusion in context, non assertive (1 mark)

b) Do test at 5% level for increased mean (11 marks)
mean = sigma(x)/n = 373/60 = 6.217 (1 mark)
s = sqrt((sigma(x^2)-(sigma(x)^2)/n)/n-1)
= sqrt (2498-(373^2/60)/59)
= sqrt(3.037) (1 mark for method of finding sample s)
=1.7427 (1 mark for explicitely giving the value of s or s^2)

H0: mean = 5.64
H1: mean > 5.64 (1 mark for both correct)
When mean = the mean level of radioactivity in limpets for the population (1 mark for defining this)

Test statistic = (6.217-5.64)/(1.7427/sqrt(60) (1 mark for dividing s by sqrt(60))
= 2.5647 (1 mark for correct value)
5% one tail critical value = 1.645 (1 mark)

Since 2.5647 > 1.645 result is significant (1 mark)

Reject H0, accept H1 (1 mark)
Conclusion, non assertive, in context (1 mark)
Phew that took SO LONG Please let me know of any errors I made Would appreciate the reps

Let me know how you think you did! I think I got anywhere from 69/72 to 72/72 based on memory of my answers not being perfect
Appreciate this. Luckily that all looks really familiar! I remember the hypothesis conclusion ratio being 2:1 (as you've done) but I can't seem to remember what order they were in or whether it was 2 rejects or 2 accepts. My numbers are the same though so I guess I must have done it the same as you.

Have a rep.

Edit: Ah yes reading the conclusions it's all flowing back. Idk why I suffer such brain-death when I leave an exam.
12. (Original post by KingAweT)
Does anyone have the m1 june 15 paper for revision purposes?
Here you go. Sorry if a bit last minute and for being a bit of a bad copy but better than nothing!

Do you know if you have to underline vectors? (eg i and j) Will we loose marks if we dont??
Attached Images
13. June 2015 M1.compressed.pdf (911.2 KB, 48 views)
14. (Original post by ComputerMaths97)
Here is my unofficial S2 MEI OCR June 2016 Mark Scheme

Spoiler:
Show

- This is me comparing the questions to previous years, and using their mark scheme. So if you disagree, I may have made a calculation error. Just let me know if you think anything should be disputed, but all methods are from previous exam boards.
- This is unofficial (yeah some people will think this is the real thing)
- If you're cheating for your mock (to you lot seeing this in 2017), remember that you gain nothing from cheating, and this is not official. S2 is mainly about learning past paper material, so you're only making it harder for yourself by not revising properly for your mock. But who needs a lecture from me ay
- But other than that, enjoy (Be sure to rep if you agree/ appreciate the couple hours I spend making this )
S2 June 2016 Unofficial Mark Scheme:
Spoiler:
Show

Q1)
i) Scatter diagram (3 marks)
Linear scales and axis labelled "Quantity of fertiliser kg/hectare" and "Percentage of population living in rural areas (%)" or similarly suitable = (1 mark)
All 11 points plotted correctly = (2 marks)
(Only 9 (or more) points plotted correctly then 1 mark instead of the 2 from above)
Note: Does not matter which axis is which variable

ii) Why pmcc test might not be valid (2 marks)
The scatter diagram does not appear to be roughly elliptical = (1 mark)
So the population may not have a bivariate normal distribution = (1 mark)

iii) Find spearmans rank correlation coefficient (5 marks)
x rank: 4 , 1 , 6 , 5 , 11 , 8 , 7 , 2 , 10 , 9 , 3
y rank: 8 , 6 , 3 , 7 , 1 , 5 , 4 , 11 , 2 , 9 , 10 (these two rows correct = 1 mark)
d^2....: 16,25,9,4,100,9,9,81,64,0,49 (row correct = 1 mark)

sum of d^2 = 366 (correct answer only = 1 mark)

r = 1 - [6 x sigma(d^2)]/n(n^2 -1)
= 1 - 6(366)/11(120)
= 1- 2196/1320 (method = 1 mark)
= -0.6636 (answer = 1 mark - allowed to 2.d.p or better)

iv) Do (negative association) test at 5% level (6 marks)
H0: No association between the quantity of fertiliser used and the percentage of the population who live in rural areas in the population of countries
H1: Some association ...

(1 mark for both hypothesis)
(1 mark for both in context)
(1 mark for "in the population of countries"

Critical value = 0.7091 (1 mark)
Since |-0.6636| < 0.7091, result not significant, therefore accept H0 (1 mark)
Conclusion in context + non assertive i.e There is insufficient evidence to suggest that there is some association between....

v) Explain the term "1% significance level" (1 mark)
The probability of rejecting the null hypothesis when it is in fact true is 1%/0.01 (1 mark)

vi) State modelling distribution assumptions for spearmans test to be valid (1 mark)
*This has never come up in an exam before. After hours of research, it's clear it's one of two answers, I'm sorry I'm unable to give a certain answer*
(Answer option 1) There are none, since spearmans rcc is calculated using rank not actual values and is therefore not concerned by actual distribution of values, just their rank
(Answer option 2) For the spearmans test to work, the underlying relationship must be monotonic.

The only reason it's not certainly number 2) is because this phrase only exists in one mei document that's only available to teachers (not in textbook, on integral, nor in past papers) and I have quoted it. Notice is says "relationship", which raises concerns to me. My teacher had never heard of monotonic so I can't be too sure.

2)
i) Define 'independently' and 'randomly' in context (2 marks)
Independently means that the occurence of one mutation does not affect the probability of another mutation occuring (1 mark)
Randomly means that the mutations are unpredictable (1 mark)

ii) Find P(X=1) (2 marks)
Binomial (20,0.012)
Therefore P(X=1) = 20C1 x 0.012 x (0.988)^19 (1 mark)
= 0.1908 (1 mark)

iii) Why is Poisson suitable as approximation (2 marks)
Since it is binomially distributed:
n is large (1 mark)
p is small (1 mark)

iv)
A) Find P(X=2) (3 marks)
lambda = np = 500 x 0.012 = 6 (1 mark)
P(X=2) = [e^-6 x 6^2]/2 (1 mark)
= 0.0446 (1 mark)

B) Find P(X>=2) (2 marks)
P(X>=2) = 1-P(X<=1) (1 mark)
=1 - 0.0174
= 0.9826 (1 mark)

v) Use suitable approximating distribution to find P(X >= 650) (5 marks)
mean = np = 50000x0.012 = 600 (1 mark)
variance = npq = 50000x0.012x0.988 = 592.8 (1 mark)
P(X >= 650) = P(Z > (649.5-600)/sqrt(592.8)) (1 mark for continuity correction)
= P(Z > 2.033) = 1 - phi(2.033) (1 mark for this CORRECT ONLY)
= 1 - 0.9789
= 0.0211 (1 mark)

3)
i) Find P(50000 < X < 55000) (3 marks)
= P([50000-50600]/3400 < Z < [55000 - 50600]/3400)
= P( -0.1765 < Z < 1.2941)
= phi(1.2941) - (1-phi(0.1765)) (1 mark for correct form here)
= 0.9022 - (1 - 0.5701) (1 mark for either of these values read of tables correctly)
= 0.4723 (1 mark)

ii) Claims that at least 95% last over 45000 hours, is he correct? (3 marks)
P(X>45000) = P(Z > [45000-50600]/3400)
= P(Z > -1.6471)
= phi(1.6471) (1 mark for this correct)
= 0.9502 (1 mark)
Therefore since 0.9502 > 0.95, claim is valid (1 mark for specifying what you just calculated means)

iii) Find h such that 99.9% of them last h or more hours (3 marks)
P(X >= h) = 0.999
......
(h-50600)/3400 = - phi(0.999)
(h-50600)/3400 = -3.090 (1 mark for +/- 3.090, 1 mark for equation in this form even if incorrect)
h = 40094 hours (1 mark)

iv) P(Y < 60000) = 0.6 and P(Y > 50000) = 0.9, find mean (x) and standard deviation (s) of Y ( 5 marks)
P(Y < 60000) = 0.6
P(Z < (60000-x)/s) = 0.6
(60000 - x) / s = inverse-phi (0.6)
60000 - x = 0.2533s (1 mark for equation in this form even if incorrect)
P(Y > 50000) = 0.9
......
50000 - x = -1.282s (1 mark for equation in this form even if incorrect)
(1 mark for either inverse-phi(0.6) = 0.2533 or inverse-phi(0.9) = 1.282 correct)
s = 6513 (or 6513.4 or 6513.39) (1 mark)
x = 58350 (1 mark)

v) Draw normal curves of X and Y on same axis (4 marks)
Two normal shapes, asymptotic behaviour and horizontal axis/values correctly labelled (1 mark)
Both means explicitely stated (1 mark)
Lower max height for Y (1 mark)
Visibly greater width for Y (1 mark)
Note: Y should be to the right of X and if either of the curves are evidently not symmetrical then a mark is lost)

4)
i) Find expected frequency of female agree, and verify test statistic of 0.6831 (3 marks)
Expected frequency = (29x42)/80 = 15.225 (1 mark)
Contribution = (12-15.225)^2 / 15.225 (1 mark for attempt at (observed-expected)^2/expected)
= 0.6831 (1 mark)

ii) Do 5% sig level test of association (6 marks)
H0: No association between the sex of the student and their attitude towards mathematics
H1: Some association between .... (1 mark for both correct)
Sum of chi-squared = 5.3236 (1 mark)
2 degrees of freedom (1 mark)
Critical value at 5% level = 5.991 (1 mark)
Since 5.3236 < 5.991, results not significant, accept H0 (1 mark)
Conclusion in context, non assertive (1 mark)

b) Do test at 5% level for increased mean (11 marks)
mean = sigma(x)/n = 373/60 = 6.217 (1 mark)
s = sqrt((sigma(x^2)-(sigma(x)^2)/n)/n-1)
= sqrt (2498-(373^2/60)/59)
= sqrt(3.037) (1 mark for method of finding sample s)
=1.7427 (1 mark for explicitely giving the value of s or s^2)

H0: mean = 5.64
H1: mean > 5.64 (1 mark for both correct)
When mean = the mean level of radioactivity in limpets for the population (1 mark for defining this)

Test statistic = (6.217-5.64)/(1.7427/sqrt(60) (1 mark for dividing s by sqrt(60))
= 2.5647 (1 mark for correct value)
5% one tail critical value = 1.645 (1 mark)

Since 2.5647 > 1.645 result is significant (1 mark)

Reject H0, accept H1 (1 mark)
Conclusion, non assertive, in context (1 mark)
Phew that took SO LONG Please let me know of any errors I made Would appreciate the reps

Let me know how you think you did! I think I got anywhere from 69/72 to 72/72 based on memory of my answers not being perfect
For the hypothesis test in question one, the person wanted to know if there was NEGATIVE association, so wouldnt H1 be there is negative association between percentage and fertiliser?
15. (Original post by groovy_q)
For the hypothesis test in question one, the person wanted to know if there was NEGATIVE association, so wouldnt H1 be there is negative association between percentage and fertiliser?
You are correct (I even put it in the question just forgot to put it in the hypothesis lool thanks)
16. If I get an answer in s2 thats one number off due to rounding do i lose marks
17. (Original post by 11234)
If I get an answer in s2 thats one number off due to rounding do i lose marks
1 at most.

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