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Edexcel Statistics 1 June 2016 Unofficial Mark Scheme

Here is a very rough unofficial mark scheme.

Anything in bold is what I am unsure of and still need confirmation. I have missed out the explanation questions as well cant remember them.

If you know anything please let me know so I can add it in!
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1. (Bee wiggles)

PMCC = 0.99
b=?
a=?
estimate = 3.581

2. (Discrete Random Variables)
a) Making equation
b and c) q = 0.15 and p = 0.175
d) E(R) = 0.45
f) Couldn't do the last part

3. (Seatrop question)

s.d = 2.51 (I used the equation that represented the sample, not the population, so might be slightly different)
Sxy = -133.9
PMCC = -0.7

4. (Venn diagram question?)
a) 0
b) u = 0.03 t = 0.22
d) P(D|RnB) = 0.45 P(D|RnB' :wink:

= 0.405
f) 16 students

5. (Baby question)

a) Width = 0.5 Height = 17
b) Median = 3.47
c) Mean = 3.43
d) Standard Distribution part i can't remember

6. (Normal Distribution)
a) ?
b) Maximum time was 206 minutes
c) P(W < u -30| W < u) = 0.36

(edited 7 years ago)

Reply 1

Sels97

What was the last part's question for #2?

Reply 2

SirRaza97

Original post by Sels97

What was the last part's question for #2?

It was the one where you had to find the probability that sarah wins or rebecca wins

Reply 3

Sels97

Original post by SirRaza97

It was the one where you had to find the probability that sarah wins or rebecca wins

IIRC Rebecca could only win when X was 1/2 and therefore she had a 0.2 chance of winning, whereas Sarah had a 0.8 chance.

I could be wrong, but this is the conclusion I came to

Reply 4

raepatricks

with that question i'm pretty sure there was the opportunity that they could draw too instead of winning/losing I think I got 0.65 for sarah and 0.2 for rebecca

Reply 5

M3WIZARD

4 D) P(D|RnB) = 0.45 P(D|RnB') = 15/37
E) 33 People

Reply 6

SirRaza97

Original post by M3WIZARD

4 D) P(D|RnB) = 0.45 P(D|RnB' :wink:

= 15/37
E) 33 People

15/37 is the same as 0.405 just rounded. How did you get 33?

Reply 7

alexhazmat

I got 16 for the people who will have dinner too

Reply 8

raepatricks

I got 33 people for dinner

Reply 9

LordStark

I got 33. Also I don't think that the dimensions for the bar on the histogram is right? I didn't get that, but I could be wrong - always struggle with histograms.

Reply 10

kieton123

Original post by LordStark

I got 33. Also I don't think that the dimensions for the bar on the histogram is right? I didn't get that, but I could be wrong - always struggle with histograms.

Mate I got 33 aswell. Always encouraging to hear that someone else got a similar answer. 😊

Reply 11

Jorjio

For 2f), I got that the probability Sarah wins is 0.475 and the probability Rebecca wins is 0.375.

From what i remember from my working out, Sarah can only win when X=3/2 and 2 so thats (0.3+0.175) and Rebecca can only win when X=-2 and 1/2 (0.175+0.2). There was a probability of drawing when X=-1 (0.15).

I could be wrong though. Did anyone else get this?

Reply 12

greentron6

Original post by Jorjio

I got the same!

Reply 13

itrytohard

1.) one question was which was the response variable, i said the number of bee wiggles is that right?
Also got something like w = 0.722 + 0.0142d for the regression line

4) last part i got 33 dinners, and i think thats right. you had to use the ratio of how many people ordered dessert to how many didn't from each of the two groups out of 77. There was a really similar question on one of the recent IAL papers

5f.) I said:
The mean and median are very similar = data is fairly symmetrical = good idea to use n.dis.
But worked out in part (d) that 25% of the babies weigh less than 3kg. Whereas the table showed that only 9/50 babies weighed less than 2 (=18%). Therefore Shyam should refine her model to make it more accurate.

6a.) what proportion finished in ?over/under? 300 minutes: 6.68%

(edited 7 years ago)