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

1. What question was 2.35 for? What value was it for?
2. (Original post by Patrick Gekko)
also wondering this.
what did you get for p and q, out of interest?
P 0.1 and Q 0.2 or the otherway around, can't remember
3. (Original post by zandneger)
My phone won't let me old half of your reply but here's why it was 0.45 and 15/37

The probability of D given BnR is the proportion of BnRnD in BnR. Since from the diagram we can see BnRnD= 0.27 and BnR = 0.6, the answer is 0.27/0.6 which gives 0.45

For probability of D given RnB' you do the same again. The total of RnB' is 0.15+0.22 = 0.37
The portion of Rn'B which is also D is 0.15, hence the answer is 0.15/0.37 which is 15/37. This is definitely right because in the next question you use the fraction to work out that 15 out of 37 people have dinner at the hotel when they order a room but no breakfast.

EDIT; I've spotted a couple of mistakes in your working. Firstly, P(Dn(RnB)) = 0.27, not 0.42, as it is the area in which all three circles intersect. Secondly, P(RnB) is 0.6, not 0.97. This is because RnB means the area inside B which is also inside R, which happens to be all of it (0.6). You've taken only the probability of R by the looks of it.
Thankyou, that mishap cost me around 2-4 marks
4. So this is my take on how many people will have dinner.
Basically i figured that some are in a room with no breakfast and some are in a room with breakfast, but nonetheless they are all in a room. So the probability we are looking for is being in a room and having dinner which is P(RibtersectionD) which comes out to be 0.42. We then multiply by 77 people to get 32.3 which is 32 people. Im not saying this is the correct method its just my take on it.

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5. (Original post by hbaig27)
P 0.1 and Q 0.2 or the otherway around, can't remember
I got 0.15 and 0.175
6. Was the median to question 4) 3.47 (basically 3.5) and was the answer about using your answers b, c and d to question 4 e) she couldn't use a normal distribution because there's a negative skew as the median is greater than the mean. To use normal distribution, there should be a symmetrical skew?

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7. (Original post by alexesl17)
So this is my take on how many people will have dinner.
Basically i figured that some are in a room with no breakfast and some are in a room with breakfast, but nonetheless they are all in a room. So the probability we are looking for is being in a room and having dinner which is P(RibtersectionD) which comes out to be 0.42. We then multiply by 77 people to get 32.3 which is 32 people. Im not saying this is the correct method its just my take on it.

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This is exactly what I got although it was kinda a guess. How did thr pmcc change when the extra balue was added?
8. Hey guys I've got my answers here:

Predicting 58 is an A and not much lower because it wasn't that bad of a paper according to most people but then again 30% found it hard. Theory questions threw me off a bit though.

Q1. 11 total i think
a) was Sdd or something (3)
b) wiggles is the response variable as it changes depending distance (1)
c) r = 0.722 + 0.014d
d) d= 350 sub in so w = 5.6 ish?

Q2 Sarah one
a) equation is 2p + q = 0.5
b) a= 0.175 b= 0.15
c) idk
d) E(R) = 0.45?
e) Sarah wins = 0.475 Rebecca wins= 0.375

Q3 with frequencies?
a) heigh of bar and width igot something like 34cm and 1cm?
according to other its 17cm height and some other width
b) show that Syy is … (1) didn’t do this right lol
c) PMCC
d) SD
e) What were the other questions?

Q4 Hotel one
a) 0
b) t =0.03
c) u = 0.22
d) 0.45 and 15/37
e) diners 16? Some people say 33

Q5
a) median 3.47
b) SD = 0.689?
c) mean proof 3.43
d) Probability baby wasunder 3kg something like 0.2546
e) explain the use of the model? (2)
I said that although the range chosen was appropriate as mean was 3.43 for both and SD was close to the calculated one, there is negative skew as median>mean so not good to use this model as mean and SD are affected by skew.
f) effect of adding baby weight of 3.43 on (3)
i) mean – I said stays the same because I sneakily calculated it , because the frequency becomes 51and value is within range or something
ii) SD – "Others said it stayed thesame, but I wrote that it decreases, because now the average distance from themean is slightly less (as n has increased by 1 but deviance hasn't changed)"

Q6.
a) 0.0668 men took over 300 min to run
b) needs to run below 206.33 minutes to be in 20% of people who finish first
c)the mew one0.18/0.5 = 0.36 (3)

Let me know what I'm missing and I've done wrong as well as the marks for each if you remember

Q1:2,1,2,4,2
Q2: 1,3,2,2,3,4
Q3: 3,1,2,2,2
Q4: 1,4,2,4,2
Q5: 3,2,3,3,3,3
Q6: 3,3,3
9. (Original post by Exavior)
This is exactly what I got although it was kinda a guess. How did thr pmcc change when the extra balue was added?
it increased ( became less negative, closer to 0) because the double increase in both values indicated a positive correlation.
10. (Original post by Florent venhari)
Was the median to question 4) 3.47 (basically 3.5) and was the answer about using your answers b, c and d to question 4 e) she couldn't use a normal distribution because there's a negative skew as the median is greater than the mean. To use normal distribution, there should be a symmetrical skew?

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I said the median and mean were so close that there was just a slight skew, because it was so small it could be seen as symmetrical and normal distribution is okay

I've seen them accept both justified sides
11. I wrote 32 dinners as instead of doing (40*0.45)+(37*15/37) = 33
I did (40*0.45)+(37*0.405) (multiplied by the decimal to 3dp) = 32.9 so wrote 32
How many marks will I receive? It was out of 2
12. (Original post by alexesl17)
they all seem pretty correct, the only difference to me was that i said the guy with the normal distribution for the babies was fine to use that equation since the mean and median of the data were pretty close indicating a symmetrical distribution
While this may be a valid answer, the other parts of the question indicated that he was wrong. The question specifically asked for you to use your answers from b), c) and d) to come to a conclusion. b) showed that the median was not equal to the mean, c) showed that the standard deviation was not 0.65 and d) showed that the proportion of babies with weight below 3kg was not the same as the table. I think you'd still get the marks if you've given a valid explanation, just that the question said to take more than just the averages into account
13. (Original post by X_IDE_sidf)
I got 0.15 and 0.175
Yep me too
14. (Original post by X_IDE_sidf)
I got 0.15 and 0.175
same
15. (Original post by alexesl17)
So this is my take on how many people will have dinner.
Basically i figured that some are in a room with no breakfast and some are in a room with breakfast, but nonetheless they are all in a room. So the probability we are looking for is being in a room and having dinner which is P(RibtersectionD) which comes out to be 0.42. We then multiply by 77 people to get 32.3 which is 32 people. Im not saying this is the correct method its just my take on it.

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that is wrong. You time the 40 by your part d(i) and your 37 by part d(ii) and you get 33
16. Was p 0.175 and q 0.15? Something like that I didn't get 0.1 or 0.2
17. (Original post by hbaig27)
P 0.1 and Q 0.2 or the otherway around, can't remember
i got something like 0.175 for one of them
18. (Original post by Bloom77)
I got 0.36 for the last one, anyone else?

Easy paper

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Got 0.36 too.Easy paper overall.
19. Did anyone else say that the mean decreased?? I know you weren't supposed to use calculations but I did just to check my answer. The new baby was 3.43 but that means that the fx value is now (18x3.25) not (17x3.25) then divided by 51 instead of 50. In ungrouped data it would stay the same but I thought because it was grouped it'd decrease?? Think I've got it wrong!!
20. (Original post by zahraaj1)
Was p 0.175 and q 0.15? Something like that I didn't get 0.1 or 0.2
yes 0.1 and 0.2 is wrong

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