Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

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#1
Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

1. i) Not everyone has an equal chance of selection (1 mark)

ii) Assign random numbers, generate using a calculator, ignore repeats. (2 marks)

2. a = 16.9 (7 marks)

3. p = 0.0076 so reject H0 (7 marks)

4. i) p = 0.0359 so accept H0 (11 marks)

ii) CLT is used when giving the distribution of x bar (1 mark)

5. i) 0.143 (4 marks)

ii) N = 90 (6 marks)

6. i) Independent, random, etc. (2 marks)

ii) 0.476 (2 marks)

iii) r=5 (4 marks)

iv) 0.0375 (5 marks)

7. i) (3 marks)

ii) a) Show ... (3 marks)

ii) b) q = (3 marks)

8. i) Condition nq > 5 not satisfied (1 mark)

ii) Show P(X=60) < 0.05 but P(X>=59) > 0.05 (4 marks)

iii) p =0.05 and Type I error = 0.0461 (2 marks)

iv) p > 0.985 (4 marks)
0
3 years ago
#2
(Original post by Mr M)
Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

1. i) Not everyone has an equal chance of selection (1 mark)

ii) Assign random numbers, generate using a calculator, ignore repeats. (2 marks)

2. a = 16.9 (7 marks)

3. p = 0.0076 so reject H0 (7 marks)

4. i) p = 0.0359 so accept H0 (11 marks)

ii) CLT is used when giving the distribution of x bar (1 mark)

5. i) 0.143 (4 marks)

ii) N = 90 (6 marks)

6. i) Independent, random, etc. (2 marks)

ii) 0.476 (2 marks)

iii) r=5 (4 marks)

iv) 0.0375 (5 marks)

7. i) (3 marks)

ii) a) Show ... (3 marks)

ii) b) q = (3 marks)

8. i) Condition nq > 5 not satisfied (1 mark)

ii) Show P(X=60) < 0.05 but P(X>=59) > 0.05 (4 marks)

iii) p =0.05 and Type I error = 0.0461 (2 marks)

iv) p > 0.985 (4 marks)
For question 4, you could have also done it using the z value method right?
0
3 years ago
#3
(Original post by Mr M)
Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

1. i) Not everyone has an equal chance of selection (1 mark)

ii) Assign random numbers, generate using a calculator, ignore repeats. (2 marks)

2. a = 16.9 (7 marks)

3. p = 0.0076 so reject H0 (7 marks)

4. i) p = 0.0359 so accept H0 (11 marks)

ii) CLT is used when giving the distribution of x bar (1 mark)

5. i) 0.143 (4 marks)

ii) N = 90 (6 marks)

6. i) Independent, random, etc. (2 marks)

ii) 0.476 (2 marks)

iii) r=5 (4 marks)

iv) 0.0375 (5 marks)

7. i) (3 marks)

ii) a) Show ... (3 marks)

ii) b) q = (3 marks)

8. i) Condition nq > 5 not satisfied (1 mark)

ii) Show P(X=60) < 0.05 but P(X>=59) > 0.05 (4 marks)

iii) p =0.05 and Type I error = 0.0461 (2 marks)

iv) p > 0.985 (4 marks)
For question 7(i) will it still get all the marks if I put 2.13 instead of 32/15

Thanks for this.
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#4
(Original post by SGHD26716)
For question 4, you could have also done it using the z value method right?
Yes.
0
#5
(Original post by SGHD26716)
For question 7(i) will it still get all the marks if I put 2.13 instead of 32/15

Thanks for this.
Yes.
0
3 years ago
#6
(Original post by Mr M)
Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

1. i) Not everyone has an equal chance of selection (1 mark)

ii) Assign random numbers, generate using a calculator, ignore repeats. (2 marks)

2. a = 16.9 (7 marks)

3. p = 0.0076 so reject H0 (7 marks)

4. i) p = 0.0359 so accept H0 (11 marks)

ii) CLT is used when giving the distribution of x bar (1 mark)

5. i) 0.143 (4 marks)

ii) N = 90 (6 marks)

6. i) Independent, random, etc. (2 marks)

ii) 0.476 (2 marks)

iii) r=5 (4 marks)

iv) 0.0375 (5 marks)

7. i) (3 marks)

ii) a) Show ... (3 marks)

ii) b) q = (3 marks)

8. i) Condition nq > 5 not satisfied (1 mark)

ii) Show P(X=60) < 0.05 but P(X>=59) > 0.05 (4 marks)

iii) p =0.05 and Type I error = 0.0461 (2 marks)

iv) p > 0.985 (4 marks)

Hi what do you think the a boundary will be for this paper. I really need a 90 ! but i think I got 67/72 do you think that will be enough?
0
3 years ago
#7
(Original post by harpoon101)
Hi what do you think the a boundary will be for this paper. I really need a 90 ! but i think I got 67/72 do you think that will be enough?
For an A* the ums needs to be an average on 90 across 3 modules so you don't need 90. However 67 should definitely be 90 ums so I wouldn't worry. At worse you might end up with 88 ums.
0
3 years ago
#8
(Original post by Mr M)
Mr K2's OCR (not OCR MEI) Statistics 2 Answers June 2017

1. i) Not everyone has an equal chance of selection (1 mark)

ii) Assign random numbers, generate using a calculator, ignore repeats. (2 marks)

2. a = 16.9 (7 marks)

3. p = 0.0076 so reject H0 (7 marks)

4. i) p = 0.0359 so accept H0 (11 marks)

ii) CLT is used when giving the distribution of x bar (1 mark)

5. i) 0.143 (4 marks)

ii) N = 90 (6 marks)

6. i) Independent, random, etc. (2 marks)

ii) 0.476 (2 marks)

iii) r=5 (4 marks)

iv) 0.0375 (5 marks)

7. i) (3 marks)

ii) a) Show ... (3 marks)

ii) b) q = (3 marks)

8. i) Condition nq > 5 not satisfied (1 mark)

ii) Show P(X=60) < 0.05 but P(X>=59) > 0.05 (4 marks)

iii) p =0.05 and Type I error = 0.0461 (2 marks)

iv) p > 0.985 (4 marks)
Just wondering for 5)ii if i got 89 not 90 not due to a rounding mistake but to a incorrect continuity correction on how many marks do you think would I lose?
0
3 years ago
#9
How many marks would I lose if I used a poisson approximation (using 0.05 for p) for the whole of question 8? (I think for part 4 I even got the same inequality! And for part 2 the same applied)
0
3 years ago
#10
I'm teaching myself S1 this year instead of doing m2 ... does anyone have an OCR S1 advanced maths essentials textbook/revision guide thingy ... it has a cd and a book ... I their guide with CD for M1 and it was amazing (went from U to an A on my own!) ... I know they do it but its not available to buy presumably because this is the last year of this spec? PLEASE LET ME KNOW I REALLY NEED ONE and would love to take it off your hands if you're finished
0
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