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Edexcel s1 June 2016 unofficial mark scheme

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Not sure for 2(f) I got E(R) the same, but there was one of which they drew which had the probability of 0.2/0.3 Therefore P(1/X>X)+P(X>1/X) = 0.8/0.7. I think I got 0.625 and 0.175 not sure though
Reply 21
Avatar for SANTR
SANTR
8
Original post by veldt127
Yeah, I think you've worded that right. It was a weird question



Ah I see, I'll get on that


For some reason I didn't get the right answer, but showed the working out.
How many marks would I drop?
Original post by SANTR
Just out of interest how did you reach the answer for 2(e). What did you have to do to the given Probability Distribution table?


for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?
I saved my answers like you and got every question the same as you, I'm pretty sure they are correct! 😀


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Original post by neyjunior
for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?


Possibly, they were very close. The mark scheme might go either way; I haven't seen this kind of question enough times to comment
Reply 25
Avatar for U6145
U6145
Original post by azizadil1998
Hey everyone, I got 0.6 for 4.a (the hotel one), is it right?? I looked at the answers above it said 0 but I just guessed anyways



what was the question again i can't remember?
Original post by veldt127
Ok, I managed to get all my answers stored in my calculator at the end, this is what I got. Not 100% sure on all of them, but we can compare.

Question

Spoiler

Question 2

Spoiler

Question 3

Spoiler

Question 4

Spoiler

Question 5

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Question 6

Spoiler


How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08
Original post by Einstein101
How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08


If you use the equation for independent events:

P(A ∩ B) = P(A) * P(B)

You get t = 0.03 and u = 0.22.
Original post by Einstein101
How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08


You were told they were independent so P(AnB)= P(A) x P(B) then rearrange. I almost forgot to add the value at the bottom of the circle, maybe you did that?
Original post by Einstein101
How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08


P (BnD)= P(B) x P(D) because they are independent.

P(D) was 0.45 I think and then you subtract everything inside D from 0.45 giving you T to be 0.03.

And then to find U, we subtract everything else from 1.
Original post by ShatnersBassoon
But the new value was exactly the same as the mean, so the SD (average distance from mean) can't increase. Others said it stayed the same, but I wrote that it decreases, because now the average distance from the mean is slightly less (as n has increased by 1 but deviance hasn't changed).


Im pretty sure i got that wrong but ill add your answer to it.
http://www.thestudentroom.co.uk/showthread.php?t=4166725
Original post by SANTR
Height for the bar is 17cm and the width is 0.5cm.


E(R) should be 2.5. Because 1/.4 is 2.5
Original post by SANTR
Height for the bar is 17cm and the width is 0.5cm.



I got the same
Original post by neyjunior
for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?
The question was 3 marks. In the worst case, no, you'd get 0/3 (1 mark for saying it's a bad model; another for saying mean =/= median; a third for comparing d to the actual answer).

You didn't compare your answer to part (d) to anything, but the question explicitly told you to, so you've definitely lost a mark there. I'd give you 1/3 but maybe if they're generous you'd get 2/3.
Original post by Florent venhari
I got these wrong:
2 f ii)
3 e) maybe I got one mark
4 d i)
4 e)
5 e) lost 1-2 marks
5 f ii)
6 c) maybe I got one mark for getting 0.5 as one of the probabilities.

Oh wells :smile:

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Im super super confused with question 6c, i had NO idea it completely threw me off, could anyone explain?
Original post by ShatnersBassoon
But the new value was exactly the same as the mean, so the SD (average distance from mean) can't increase. Others said it stayed the same, but I wrote that it decreases, because now the average distance from the mean is slightly less (as n has increased by 1 but deviance hasn't changed).


I'm pretty sure I got it wrong haha but ill add your answer and others' so people can discuss it.
Reply 36
Avatar for haes
haes
6
Original post by veldt127
Ok, I managed to get all my answers stored in my calculator at the end, this is what I got. Not 100% sure on all of them, but we can compare.

Question 1

Spoiler

Question 2

Spoiler

Question 3

Spoiler

Question 4

Spoiler

Question 5

Spoiler

Question 6

Spoiler



I said all of these except 5e! I put yes because the mean and median are similar but I didn't even understand the question lol
Original post by pranto96
E(R) should be 2.5. Because 1/.4 is 2.5


I thought that initially but worked through:

13441821_10207178478156340_594191193_o.jpg
Original post by veldt127
I thought that initially but worked through:

13441821_10207178478156340_594191193_o.jpg

I totally get your point. But why don't we get the right answer if we do 1/.4?
Original post by pranto96
I totally get your point. But why don't we get the right answer if we do 1/.4?


Well, E(x) is the sum of x over n, right?

E(1/x) is the sum of 1/x over n.

(p(X = x)( (Σ1/x) )/n is not the same as (p(X = x)(1/( (Σx)/n ))
(edited 7 years ago)

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