OCR MEI Statistics 1 (S1)

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There is a complex formula used to calculate UMS, for those who have done S2, like myself, it's done by Normal distribution, taking mean and standard deviation of all of the scores into account, then from this they work out the grade boundaries, so if people found it as hard as they said they did, I'm going for around 54/72 for an A, but it changes, C3 Jan 2012 was the hardest exam I've ever done, and the grade boundary was still 55, even though around half of my year failed the exam (got a U), so it may likely be 56 or something.... 90 UMS is going to be APPROXIMATELY 6/7 marks above the grade boundary, it's around 7 marks per grade boundary normally. So if the A boundary is 54, you'll need around 61/62 for 90 UMS and about 68/72 for 100 UMS

Thanks very much :smile:

Reply 181

mailman93

I think if it was harder than June 2008- which had a reaally low grade A boundary- then this may have an even lower one.

You can google variations of Raw to Ums converter mei ocr statisics 1 and whatever paper year and month. Hope that is what you mean- if it is a conversion of what you think you got in this morning's paper- you can only find out when MEI finds out how the whole country did and adjusts the marks "so that Avisha's 55/72 in january is the same achievement as Taniqua's 52/72 in june after the june paper was considered harder"

I don't know whether I hate the UMS system- the worst think is not recieving a damn mark or a proper grade- just an obscure lower case letter you have to look for!!

anyway, I DON'T CARE anymore- it's sunny

:smile:

seees yas

Reply 182

p_5

Original post by mailman93

After Jack's beasting- I think that it will be lower than 56, mainly because of how much stuff there was on binomial probability stuff which is tricky and easier for stats 2 people. But other than being a trickier paper than a lot of past papers -including June 2008 with the 53/72 A grade boundry- there was a lot of time consuming crap to do!! The hypothesis test AND drawing the histogram AND a line graph all take a lot of time- especially if you're not sure what you're doing!

SO- could be one of the lowest? Or is that wishful thinking?? I genuinely think 51/52/53

I dont think it would be that low to be honest...i think something like 55/56+
how much was the one last question out of?
And you know for the question with the question where they asked about the 2.5 million car - didnt you have to times 781500 by 1000? to get £781250000.

Reply 183

username537823

Original post by p_5

yes you did!

Reply 184

AndrewCoppola

For the last question I got 0.023 or something with 18 combinations, anyone else? The 10 combination idea seems more popular :frown:

Reply 185

Jack0234

Original post by p_5

I'm inclined to agree I think it'll be around 56. The last question for getting 3 heads was out of 4 marks and yes I agree I got the same value for the car question.

Reply 186

Alotties

My teacher never taught us any like annotation things to put before our probabilities or any probability formulas (cept standard deviation) but I've seen a lot of people write them on here!
Will this lose me marks for not writing them? Because I've honestly never seen them before... :confused:

Reply 187

amg_22

Anybody else get 0.009693 for the last question

(6 and 3 may be other way round can't remember exactly lol)

Reply 188

username537823

Original post by amg_22

Anybody else get 0.009693 for the last question

(6 and 3 may be other way round can't remember exactly lol)

yes that is the right answer, which I also got... no more on that question that is the right answer

Reply 189

thomasdevlin

made a mistake on question 1, just put (30/50)^3 :frown:

did the question actually say that she didn't replace the plants? if it didn't say that then technically i could still be right, can't remember though

Reply 190

amg_22

Original post by sreddy17

yes that is the right answer, which I also got... no more on that question that is the right answer

I'm guessing I'm not the first to ask this then lol!

Reply 191

Deceptively

Original post by thomasdevlin

made a mistake on question 1, just put (30/50)^3 :frown:

did the question actually say that she didn't replace the plants? if it didn't say that then technically i could still be right, can't remember though

Nah, I did what you did (just cubed it) but because it gave specific numbers I think you had to decrease it by one after all... if it was just probabilities (ie assuming an infinite number of flowers to choose from) then it would just be cubing, but that makes sense now that it's without replacement. Darn, I guess I did worse than I thought.

Even so, I'll use the answers from you guys for the ones I did wrong and try to get cracking on a mark scheme... I'll start with just the correct method/answers really quickly and then get some suggestions for A1, M1 etc. Though I've never done one of these before so I hope it turns out alright! Feel free to tell me when I've got something wrong, though!

Reply 192

mailman93

Original post by p_5

last question part was out of 4 marks

and yes I got that ^^^

Reply 193

PhoenixxX

How many marks would i get for adding instead of multiplying the permutations?

Reply 194

Deceptively

Stats 1 2012 Mark Scheme
* The M1, A1 etc. are only suggestions. I can't remember all the mark allocations so please do tell me what they were so this can be more accurate. Also please tell me where I've made a mistake and I'll fix it. I've put a brief summary of the question to remind you which questions they are.

1) Karen buys three bulbs out of 50.
i) Find P(All 3 are blue)
30/50 * 29/49 * 28/48 = 29/140 (= 0.20714…)
M2 for 3 correct fractions. M1 for 2 correct.
A1 CAO
ii) Find P(At least one of each colour)
1 - P(all the same colour) (M1)
= 1 - 29/140 - 20/50 * 19/49 * 18/48 (M1)
= 0.7381... (A1)

2) An exam paper has 5 questions in section A and 9 in section B.
i) How many ways can you choose 3 from A and 3 from B?
5C3 * 9C3 (M1)
= 840 (A1)
ii) Find P(Choose 3 from each section) if a candidate chooses 6 questions randomly.
14C6 = 3003 (M1)
(Their 840)/(their 3003) (M1)
40/143 (=0.2797...) (A1)

3. 85% of callers are put on hold. A sample of 30 callers is taken.
i) P(29 put on hold)
Attempt at using binomial distribution (M1)
30C29 * 0.85^29 * 0.15^1 (M1)
= 0.0404 (A1)
ii) P(At least 29 put on hold)
(Their 0.0404) + 0.85^30 (M1)
= 0.0480 (A1 (FT?))
iii) 10 samples of 30 callers. Find expected number of samples in which at least 29 callers are put on hold.
(Their 0.0480)*10 (M1)
= 0.48 (A1 (FP?)- must not round to whole number)

4. 8% of the population uses a particular web browser.
i) A) P (The first person to use that browser is the 3rd one he asks)
0.92*0.92*0.08 (M1)
= 0.067712 (A1)
B) P(First person to use the browser is the 2nd or 3rd person he asks)
= 0.067712 + 0.92*0.08 (M1)
= 0.141312 (A1)
ii) P(At least one of the first 20 uses the browser)
1 - P(None of the twenty uses the browser) (M1)
= 1 - 0.92^20
= 0.8113 (A1)

5. Hypothesis test at 5% significance; 5% of bikes are faulty.
p = Probability that a bike is faulty (M1)
H0: p = 0.05 (M1)
H1: p > 0.05 (M1)
EITHER
P (X >= a) < 0.05
1-P(X<= a-1) < 0.05
P(X<= a-1) > 0.95
Critical region is {4,5,6,7,8,9,10,11,12,13,14,15,16,17,18} (M3)
OR
Method not using critical region (M3)
(I never learnt the other method so can someone fill this in for me?)

"4 is in the critical region" (A1 comparative statement)
(There is sufficient evidence to reject H0 and accept H1...)
Conclusion in context eg. "The proportion of fault bike frames has increased" (A1)

6. Histogram. x values: 500-1000, 1000-1500, 1500-2000, 2000-3000, 3000-5000
f values respectively: 7, 22, 26, 18, 7
i) Draw histogram (I don't have any way to draw this so can anyone draw one for me?)
Frequency density values: 0.014, 0.044, 0.052, 0.018, 0.0035 (M2 all correct; -1ee)
Histogram:
A1 axes correct (linear scale starting at 0, labels)
A1 heights correct
A1 widths correct
ii) Student claims the midrange is 2750; is he likely to be correct?
He might be correct since this is between the minimum and maximum possible midrange values. (A1)
iii) Calculate mean and standard deviation. Explain why this is only an estimate.
Mean: (750*7 + 1250*22 + 1750*26 + 2500*18 + 4000*7)/80 (M1)
= 1890.625 (A1)
Standard deviation: sqrt(((750^2*7 + 1250^2*22 + 1750^2*26 + 2500^2*18 + 4000^2*7)-80*1890.625^2)/79) (M1)
= 845.54 (A1)
Only an estimate because the data are grouped. (A1)
iv) Investigate whether there are any outliers.
1890.625+2*845.54 = 3581.7 (M1)
So there are probably outliers at the upper end of the distribution (A1)
1890.6-2*845.54 = 199.5
So there are no outliers at the lower end of the distribution. (A1)
v) Vehicle duty of £1000 for engines larger than size 2000. Estimate how much money is raised from 2.5million new cars.
(18+7)/80 * 2500000 * 1000
= £781250000 (A1)
vi) Why might this estimate be too high?
A1 for any reasonable statement eg. tax evasion, new engines will be made smaller to avoid the duty, etc.

7. 5 coins; one is biased with P(Heads)=0.6. X=number of heads.
i) Show that P(X=0) = 0.025
0.4*0.5^4 (M1)
= 0.025 (A1 - NB answer given)
ii) Show that P(X=1) = 0.1375
0.6*0.5^4 (M1)
+ 4*0.4*0.5^4 (M2)
= 0.1375 (A1)
iii) Draw a line chart.
A2 Line heights correct -1ee
iv) Comment on skewness
Negative skewness (A1)
v) Find E(X) and Var(X)
E(X) = 0*0.025 + 1*0.1375 + 2*0.3 + 3*0.325 + 4*0.175 + 5*0.0375 (M1)
= 2.6 (A1)
Var(X) = 0^2*0.025 + 1^2*0.1375 + 2^2*0.3 + 3^2*0.325 + 4^2*0.175 + 5^2*0.0375 - E(X)^2 (M1)
= 8 - 2.6^2 (M1)
= 1.24 (A1)

vi) 6*P(0,1,2) + 3*P(0,0,3) + P(1,1,1) (M1)
= 6*0.025*0.1375*0.3+3*0.025*0.025*0.325+0.1375^3 (M2)
= 0.009396 (A1)

Suggested grade boundary for A - 55/72 (Based on comments on this thread).

(edited 11 years ago)

Reply 195

nju

I hope the grade boundary is not 60 again I really want it to be about 55 but that is just wishful thinking

(edited 11 years ago)

Reply 196

helpme12345

For the last question I only had a couple of minutes left so was rushing and made a really silly mistake :frown:

I did 3P(3, 0, 0) + 3P(1, 2, 0) I forgot P(1, 1, 1) and mistakenly put a 3 infront of P(1, 2, 0) instead of a 4
It was 4 mark so would I get any marks at all?? The 3P(3, 0, 0) was right, and P(1, 2, 0) but not the number infront...

Also for the 2nd "show that" question of the Q7 show P(x=1) =??? I couldn't figure out how many arrangements of "1 heads" you could with the biased coin giving a tails.
I know this is wrong, but I did 5(0.6 x 0.5^4) + (0.4 x 0.5^4) I panicked and didn't finish, I know it was wrong since it didn't show it was equal to whatever they gave, and I just stuck the 5 there out of panic... would I get any marks for this?? :/
(I multiplied by 5, because i was thinking HTTTT OR THTTT OR TTHTT OR TTTHT OR TTTTH but knew it was wrong since I couldnt get the answer, but left it there since I couldnt think of anything else :s)

(edited 11 years ago)

Reply 197

KageKitsune

Original post by AndrewCoppola

For the last question I got 0.023 or something with 18 combinations, anyone else? The 10 combination idea seems more popular :frown:

I did that too.. Any teachers on here to give the correct answer please? ^^

Reply 198

Axion

HORRIBLE PAPER imo :facepalm:

Reply 199

stefanocattaneo

Original post by Deceptively

Stats 1 2012 Mark Scheme
There, that's all of section A, please do correct where I've made mistakes (especially with the number of marks per question - I made up a lot of those).