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Edexcel S1 - 15th June, 2016 [Exam Discussion] Watch

1. 0.5cm^2 was the area that represented 1 person.
since the class had 17 people, we can find the area by multiplying 0.5 by 17, which equalled 8.5cm^2
and the width for that question was 0.5cm
therefore, to find height we divide 8.5 by 0.5, which equalled 17
2. i got 40 something people to eat dinner what about everyone else?
3. (Original post by WashingPowder)
1. 9.44 , y= 0.722 + 0.0142x (Not with y and x , but w/e they gave you) , 5.692 ,
2 or 3. 0.15 , 0.175 , 0.65, -0.754?
4. 0.03 (Probability of D but nothing else?), 0.22 (Just Room no dinner/breakfast), 0.45 (The p(D | B overlap R)), 0.405 (p(D | B overlap R' )), 33 (No of Dinners? or w/e it was)
5. 17cm (height) , 3.39 (Median) , 3.43 (mean), 0.68 (Standard Deviation), 0.2546 (The probability).
6. 0.0668 (Probability of W > 240??) , 206.33m (Minimum distance Nathaniel had to run to be in top 20%) , 0.82 , 0.36 (The P ( z < mu - 40 | z< mu)) or w/e it was

Theres my answers, See if you got any the same
How did you work out the number of diners?
4. (Original post by mbh16)
i got 40 something people to eat dinner what about everyone else?
I dont hink anyone got coherent answers on this one lol
5. (Original post by WashingPowder)
1. 9.44 , y= 0.722 + 0.0142x (Not with y and x , but w/e they gave you) , 5.692 ,
2 or 3. 0.15 , 0.175 , 0.65, -0.754?
4. 0.03 (Probability of D but nothing else?), 0.22 (Just Room no dinner/breakfast), 0.45 (The p(D | B overlap R)), 0.405 (p(D | B overlap R' )), 33 (No of Dinners? or w/e it was)
5. 17cm (height) , 3.39 (Median) , 3.43 (mean), 0.68 (Standard Deviation), 0.2546 (The probability).
6. 0.0668 (Probability of W > 240??) , 206.33m (Minimum distance Nathaniel had to run to be in top 20%) , 0.82 , 0.36 (The P ( z < mu - 40 | z< mu)) or w/e it was

Theres my answers, See if you got any the same
i remember the median to be 3.47
6. (Original post by kelvin1745)
0.5cm^2 was the area that represented 1 person.
since the class had 17 people, we can find the area by multiplying 0.5 by 17, which equalled 8.5cm^2
and the width for that question was 0.5cm
therefore, to find height we divide 8.5 by 0.5, which equalled 17
oh nice i got this too
7. (Original post by wr123)
I'm thinking more along the lines of 58/75 for an A
Last year's was harder tho?
How did you work out the number of diners?
expected =probability x number

you times the probabilities you got in the previous ones by the numbers they gave you and add them.

its the same thing as E(x)
9. The answer for the dinner q was 33!
10. (Original post by alexesl17)
i remember the median to be 3.47
i do too
11. (Original post by mbh16)
i got 40 something people to eat dinner what about everyone else?
i got 33
12. (Original post by Apolexian)
expected =probability x number

you times the probabilities you got in the previous ones by the numbers they gave you and add them.

its the same thing as E(x)
I did this and I got 33 Lol
What was the WJEC S1 exam like?
Personally i found it pretty difficult but in general the people i have talked to have had a better time but unfortunately as i said earlier hardly anyone does WJEC
14. For question 2, I got Var(X) = 2.115.
For question 6, I got 4 hours 34 minutes (or 274 mins).

(Original post by Florent venhari)
For question 6.a, i left it as the probability. Will i lose a mark for not multiplying by 100?
Absolutely not. You can never just randomly multiply anything by 100 - you could have multiplied by 100% (which is equal to one) - but the question did not specify to leave your answer as a percentage so any form (decimal, fraction, percentage, heck even a per mille) should be fine.
15. (Original post by Siddhart1998)
Last year's was harder tho?
true im thinking 60 for an A
16. [QUOTE=Florent venhari;65797623]For question 6.a, i left it as the probability. Will i lose a mark for not multiplying by 100?
No i have seen on previous mark schemes they will allow it
17. wheres the unofficial mark scheme at for edexcel s1.........
18. did anyone get 0 for one of the probability questions?

19. Its very similar to this question from the June 2014(AIL) which luckily i did last night. But you basically , for the question above you would do 27 * P(D | M n S) + 36 * P(D | M n S') to find the total that had desserts
How did you work out the number of diners?
20. (Original post by mbh16)
i got 40 something people to eat dinner what about everyone else?
yeah I got like 47

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