Mr K's OCR (not OCR MEI) S1 answers May 2012

For 14 numbers the LQ is (14+1)/4 . . which is the 3.75th value, not the 4th Sometimes the mark schemes allow for using the 4th, sometimes not

No, if it is an even set of values, you divide them into two groups (1-7), (8-14).
Then, you find the median of the lower group, which is the 4th number, 23

Mr K's OCR (not OCR MEI) S1 answers May 2012


1. (i) r = -0.709 (4 marks)

(ii) y = -0.0185x + 8.5742

Estimate £3394 (4 marks)

(iii) Correlation does not imply causation (1 mark)


2. Mean = 1.53 kg

Standard deviation = 14.7 g (6 marks)


3. (i) 22.5 (1 mark)

(ii) w = 0 and x = 0 (2 marks)

(iii) 40 or 40.75 (2 marks)

(iv) Gives a visual display of distribution OR original data preserved (1 mark)

(v) Shows spread/IQR/median more clearly (1 mark)


4. (i) Top pair of branches Red 4/5 and Black 1/5

Bottom pair of branches Red 1 and Black 0 (2 marks)

(ii) 1/3 (3 marks)

(iii) 4/5 (2 marks)


5. (i) (a) 1 (1 mark)

(b) -1 (1 mark)

(ii) 0 (3 marks)

(iii) Completely agreed, Completely disagreed, No agreement (3 marks)


6. Profit = £110 (5 marks)


7. (i) (a) 2520 (2 marks)

(b) 720 (3 marks)

(ii) (a) 21 (1 mark)

(b) 10/11 (3 marks)


8. (i) 0.825 (2 marks)

(ii) (a) 0.265 (2 marks)

(b) 0.168 (6 marks)


9. (i) (a) 0.0590 (3 marks)

(b) 0.410 (3 marks)

(ii) (a) 0.0951 (2 marks)

(b) 0.513 (3 marks)


These answers were produced by my colleague Mr K and not by me so there is probably little point asking me about them. Hope you find them useful though!

Not even sure what you are doing there.

6P4 x 2 surely?

I taught this module, and did this paper myself today. I haven't got my answers with me, but they look correct with the exception of this one (q3) - I think the students querying this are correct. For a list of 14 numbers the median is the mean of the 7th and 8th, the LQ is the 4th (so 23 is the correct answer for (i) ) and the UQ is the 11th number. (The 'median' of the 7 above the median). That is the number marked as 3|z which can only represent 38 or 39. The gap to 46 is irrelevant.

By the way, thanks for your efforts Mr M, I know many of my students appreciate it.

I was just looking at the two situations. Say you have one vowel, then the vowel has to be at the end, so can only go in one space (the end), and the other 4 letters can go anywhere, so 4!*1 =4!. And then when you have two vowels, either of the two vowels can be at the end, so 2*4!.

Why is it 6P4?

You are forgetting the other letters. You start with 7 letters and 5 places. The A must be at the end so you put it there, you then have 4 other places which 6 letters can fill hence 6P4 (permutations as asked about arrangements). You then do the same for te other vowel E.

Resulting in 6P4 * 2

Was only worth 2 marks I think, good luck.

how do people even manage to remember their answers?! they were mostly decimals I wish i had a good memory

anyway does anyone know what you need to get for an A*? Is it an average of an A* for your A2 units and an A at As or do you need an average of an A at A2 with 90%+ on each paper? My teacher's confused!

For an A* in maths, you need an average of 90% UMS over C3 and C4 (eg 90 90, or 80 100) and then an average of an A in C1, C2, S1 and M1 (or whatever modules you're taking!).

For an A* in maths, you need an average of 90% UMS over C3 and C4 (eg 90 90, or 80 100) and then an average of an A in C1, C2, S1 and M1 (or whatever modules you're taking!).

Wait, so you only need 90% UMS in two modules?? If I did C1, C2 and S1 in my first year and got an A, and then got 90% UMS in C3 and C4, but only 80% in D1 (which I'd do in my second year), would that be an A*? Don't I need 90% in all second year modules?