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# Edexcel S1 Unofficial Markscheme - 7th June 2017 watch

1. Not totally sure if some of my answers are correct but i thought i'd share them before i forget
1.) Swt = -49 Stt = 74.8333
S ss = 50,000,000 Sst = -49000
r = -0.801
Regression line is suitable as strong negative linear correlation
b = -0.655 a= 20.0
s = 20000 - 655t
For an increase in 1 degrees sales decrease by £655

Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28 t= 0.17
Events were not independent
For the given probability 0.43 / 0.75

Measures of location Q :
Width = 4 cm height = 1.5 cm
Median : 49 mean : 49.1 SD : 10.32
Almost no skew as mean and median take similar values , or slight positive skew as mean > median
Normal distribution is suitable
80th percentile = 58.416

First discrete random variables Q :
a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

Normal distribution Q :
0.3446 for part a.)
Can only remember my answer for part c.) Was 19.75

Last discrete random variables Q :
P= 1 /3 so P(2) = 2/3
Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
Charlotte
2. Ahhhh i literally recognise about three of yours answer. Looking forward to my non existent A grade
3. For the last question I got for the probability outcomes 0,6,12,15,30
4. (Original post by ErinMei)
Not totally sure if some of my answers are correct but i thought i'd share them before i forget
1.) Swt = -49 Stt = 78.3
S ss = 50,000,000 Sst = -49000
r = -0.801
Regression line is suitable as strong negative linear correlation
b = -0.655 a= 20.0
s = 20000 - 655b
For an increase in 1 degrees sales decrease by £655

Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
Events were not independent
For the given probability 0.43 / 0.75

Measures of location Q :
Width = 4 cm height = 1.5 cm
Median : 49 mean : 49.1 SD : 10.32
Almost no skew as mean and median take similar values , or slight positive skew as mean > median
Normal distribution is suitable

First discrete random variables Q :
a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

Normal distribution Q :
Can only remember my answer for part c.) Was 19.75

Last discrete random variables Q :
P= 1 /3 so P(2) = 2/3
Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
i do not agree with your regression line , it should be s=20000-655t, but I agree with everything else, also last question, it was charlotte because she had a higher expected score ( E(X^2) is greater than E( S)
5. (Original post by ihatehannah)
i do not agree with your regression line , it should be s=20000-655t,
Woops i meant to put that 😜

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6. I got that 655 for b but I hear some people are saying 900 something. Can you remember the working out because I can't
7. Thanks! Got nearly all of the same answers - did you get Charlotte winning at the end because E(x^2) > E(S)
8. how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
9. (Original post by Chris_isaacs)
For the last question I got for the probability outcomes 0,6,12,15,30
I may be wrong but pretty sure 0 wasn't one of them
10. Did anyone else get charlotte for last part??
11. (Original post by ErinMei)
Not totally sure if some of my answers are correct but i thought i'd share them before i forget
1.) Swt = -49 Stt = 78.3
S ss = 50,000,000 Sst = -49000
r = -0.801
Regression line is suitable as strong negative linear correlation
b = -0.655 a= 20.0
s = 20000 - 655t
For an increase in 1 degrees sales decrease by £655

Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
Events were not independent
For the given probability 0.43 / 0.75

Measures of location Q :
Width = 4 cm height = 1.5 cm
Median : 49 mean : 49.1 SD : 10.32
Almost no skew as mean and median take similar values , or slight positive skew as mean > median
Normal distribution is suitable

First discrete random variables Q :
a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

Normal distribution Q :
Can only remember my answer for part c.) Was 19.75

Last discrete random variables Q :
P= 1 /3 so P(2) = 2/3
Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
Think i dropped a total of 5-6 marks unless i get method marks
12. For the yuto did you do p (median >18 | TIME > 15) = 0.5?
13. (Original post by stephaniecheung)
how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
Because they wanted it for w on t, not s on t
14. (Original post by ErinMei)
Not totally sure if some of my answers are correct but i thought i'd share them before i forget
1.) Swt = -49 Stt = 78.3
S ss = 50,000,000 Sst = -49000
r = -0.801
Regression line is suitable as strong negative linear correlation
b = -0.655 a= 20.0
s = 20000 - 655t
For an increase in 1 degrees sales decrease by £655

Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
Events were not independent
For the given probability 0.43 / 0.75

Measures of location Q :
Width = 4 cm height = 1.5 cm
Median : 49 mean : 49.1 SD : 10.32
Almost no skew as mean and median take similar values , or slight positive skew as mean > median
Normal distribution is suitable

First discrete random variables Q :
a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

Normal distribution Q :
Can only remember my answer for part c.) Was 19.75

Last discrete random variables Q :
P= 1 /3 so P(2) = 2/3
Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
I agree with everything
15. I agree with those answers
16. (Original post by stephaniecheung)
how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
It is
17. I had Sam got a higher score from E(S)> E(X^2)? maybe I just calculated it wrong
18. I got 0.3446 for 5.a) and 0.4749 5.b)
20. So I definitely know I got q1 b and q5 c wrong but I got everything else right.

That would be pretty much 0.906

Now I'm just hoping for low grade boundaries. Thanks for the answers

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