june 2003 s2 paper!!! help Watch

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bigt
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anyone know where i can find the solutions/answers for the edexcel june 2003 s2 paper. had real toruble with this and now don't know if i done it right!. i've tried the mr hughes site, but he seems to be missing the solutions for this paper!

any ideas?
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hatton02
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I have the June 2003 S2 provisional mark scheme here (EdExcel). I will post what the answers are: Sorry for any inaccuracies, I can't draw diagrams etc, but I'll try my best for you!
I will put the mark allocation at the end of each part too to help.
P.s I don't have the questions, so I hope this makes sense!

Note:
>= means "Greater than or equal to"
<= means "Less than orr equal to"
/= means "Does not equal"
ft means "follow through marks"
(*) means "final line is given on the paper"
H0 = H with subscript 0, i.e. hypothesis testing
H1 = H with subscript 1, i.e. hypothesis testing
x0 = x with a subscript of 0, used on question 7.

Question 1)
(a) A random variable; that is, a function involving no other unknown quantities. B1; B1 (2)
(b) If all possible samples are taken; then their values will form a probability distribution called the sampling distribution. B1;B1 (2)

TOTAL 4 MARKS

Question 2)
(a) lambda is large or lambda > 10. B1 (1)
(b) Y~N(30,30) .... may be implied B1
P(Y>28) = 1 - P(Y<=28.5) M1 A1
............= 1-P(Z<=(28.5-30)/root 30) M1 A1
............= 1-P(Z<=-0.273)
............= 0.607 A1 (6)

TOTAL 7 MARKS

Question 3)
(a) X~B(4,0.3) B1 B1 (2)
(b)
There is a diagram here, so I cannot draw it. On the x-axis the numbers 0, 1, 2, 3 and 4 appear. On the y-axis, it is labelled as "probability" with values up the side of 0.1, 0.2, 0.3 and 0.4
On the "0" line, it reaches up to 0.240
On the "1" line, it reaches up to 0.4116
On the "2" line, it reaches up to 0.2646
On the "3" line, it reaches up to 0.0756
On the "4" line, it reaches up to 0.0081

All probabilities correct B1
Scales and labels B1
Correct diagram B1 (3)

(c) 1 resident B1 (1)
(d) E(X) = np = 1.2 B1
Var(X) = np(1-p)
........= 4x0.3x0.7 M1
.......=0.84 A1 (3)

TOTAL 9 MARKS

Question 4)
(a) Fixed number of independent trials B1 B1 (2) (You need "fixed number" and "independent" for the 2 marks)
2 outcomes B1
Probability of success constant B1 (4) (You need "constant" for this B1 mark)
(b) P(X=5) = 2/7; P(X/=5) = 5/7 (may be implied) B1;B1 ft
P (5 on sixth throw) = (5/7)^2 x (2/7) ---- or p^n(1-p)
M1 A1 ft
.............................= 0.0531 A1 (5)

(c) P(exactly 3 fives in first eight throws) = 8C3 x (2/7)^3 x (5/7)^3
(Use of nCr needed) M1 A1 ft
..... = 0.243 A1 (3)

TOTAL 12 MARKS


Question 5)
(a) f(x) = 0.05 180<=x<=200
f(x) = 0 otherwise B1 B1

There is a graph here, with horizontal lines, between 0 and 180 (on the x-axis), between 180 and 200 (raised slightly, it is on the 0.05 mark), and then 200 upwards (it is on the x-axis).
The horizontal axis is labelled 'x' and the vertical axis is labelled f(x)

labels B1
3 parts B1

(b)(i) P(X<=183) = 3x0.05 M1
= 0.15 A1
(ii) P(X=183)=0 B1 (3)

(c) IQR =10 B1 (1)
(d) 0.05(200-x); = 0.05(x-180) x2 M1; A1
200-x=2x-360
So x=186 2/3 A1 (3)

(e) 1/3 of all cups of lemonade dispensed contains 186 2/3 ml or less
B1 B1 ft (2)
(or 2/3 of all cups of lemonade dispensed contains 186 2/3 ml or more)

TOTAL 13 MARKS

Question 6)
(a) Po(1) B1 B1

Each patient seen singly OR patients with disease seen randomly OR seen constant rate of once per week OR each patient assumed independent of the nxt B1 (3)

(b) X~Po(4) (may be implied) B1
P(X>3)=1-P(X<=3) M1
..... =1-0.4335 A1
.... = 0.5665 A1 (4)

(c) H0: lambda = 6 B1
H1: lambda < 6 B1
P(X>=2)=0.0620 alpha=0.05, implies critical region X<=1 M1 A1
0.0620 > 0.05 2 not in critical region M1

The number of patients with the disease seen by the doctor has not been reduced. A1 (6)

(d) This does not support the model as the disease will occur in outbreaks; the patients seen by the doctor are unlikely to be independent of each other/don't occur singly. B1;B1 (2)

TOTAL 15 MARKS

Question 7)
(a) Integral between 0 and -1 of k(x^2+2x+1) dx = 1 limits needed and =1 for the mark M1

[k(x^3/3 + x^2 + x)]between 0 and -1 = 1 attempt at integration M1 A1

k=3 (*) A1 (4)

(b) E(X) = integral between 0 and -1 of x f(x) dx M1
= integral between 0 and -1 of (3x^3 + 6x^2 + 3x) dx limits needed for mark A1
= [3x^4/4 + 2x^3 + 3x^2/2] between 0 and -1 integration and subtracting limits M1
= -1/4 A1 (4)

(c) Integral between x0 and -1 of (3x^3 + 6x^2 + 3x) dx = [x^3 + 3x^2 + 3x]between x0 and -1. M1
= xo^3 + 3xo^2 + 3xo + 1 A1

....... 0 .......................... x<-1
f(x)... x^3 + 3x^2+3x+1 .... -1<=x<=0
........1....................... ..... x>0

All that for B1 B1 (4)

(d) P(-0.3<X<0.3) = F(0.3)-F(-0.3) M1
...= 1-0.343 A1
...=0.657 A1 (3)

TOTAL 15 MARKS.

TOTAL FOR PAPER = 75 MARKS.

Hope this helps! Phew, finally finished typing!
Regards
"Countdown Kirk"
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Hoofbeat
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(Original post by bigt)
anyone know where i can find the solutions/answers for the edexcel june 2003 s2 paper. had real toruble with this and now don't know if i done it right!. i've tried the mr hughes site, but he seems to be missing the solutions for this paper!

any ideas?
What is this Mr Hughes site?! :confused:
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