wjec s1 jan 2007 marking scheme
DESCRIPTION
WJEC S1 January 2007 mark schemestatistics GCETRANSCRIPT
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MATHEMATICS S1
1 (a) Prob = 211
72
83
94 = or 3934 CC M1A1
(b) Prob = 215
74
85
96 = or 3936 CC M1A1
(c) Prob = 726
74
83
92 = or 39141312 CCCC M1A1A1
2 (a) 28.038.048.0)( += BAP M1 = 0.58 A1 (b) 58.01)( = BAP M1 = 0.42 A1 (c) 1.0)( = ABP B1 52.0)( =AP B1
)()A(
AP)AP(BBP = M1
= )192.0(265 A1
3 Mean = 1.0n B1 SD = 9.01.0 n B1
9.01.01.0 = nn M1 nn 09.001.0 2 = A1
n = 9 (cao) A1
4 (a) Choose 2, Prob all heads = 1/4 M1 Choose 3, Prob all heads = 1/8 Choose 4, Prob all heads = 1/16 A1
P(All heads) = 161
31
81
31
41
31 ++ M1
=487 A1
(b) P(2All heads) = 48/712/1 B1B1
=74 B1
[In (b) award B1 for each correct number]
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5 (a)(i) X is B(20, 0.35) (si) B1
P(X = 5) = 155 65.035.0520
M1
(or 0.2454 0.1182 or 0.8818 0.7546) = 0.1272 A1
(ii) P(X < 8) = 0.601 M1A1 (b)(i) Y is B(500, 0.03) and therefore approx Po(15). B1
P(Y = 10) = 0486.0!10
15.e10
15 = M1A1 (or 0.1185 0.0699 or 0.9301 0.8815) (ii) P(Y > 12) = 0.7324 M1A1 6 (a) Sum of probs = 1 so p + q = 1 0.45 = 0.55 B1 (b) E(X) = 0.3 + 2p + 0.3 + 4q + 0.25 = 2.75 M1A1 So 2p + 4q = 1.9 A1 Substituting from 1st equation into 2nd, M1 2(0.55 q) + 4q = 1.9 q = 0.4 and so p = 0.15 AG
[For candidates who simply verify that E(X) = 2.75 with the given values of p and q, award M1A1] (c) 05.025...15.043.01)( 2 +++=XE M1 = 9.45 A1 Var(X) = 9.45 275.2 = 1.8875 A1 (d) (i) E(Y) = 4 2.75 + 2 = 13 M1A1
Var(Y) = 16 1.8875 = 30.2 M1A1 (ii) P(Y < 15) = P(X < 3.25) = 0.55 M1A1
7 (a)(i) E(X) = xxxx d).(20 41
0
3 M1A1 =
1
0
65
6520
xx A1
=32 A1
(b)(i) F(x) = dyyyx
0
3 )1(20 M1
(limits not required here)
= x
yy
0
54
5420
A1
= 54 45 xx A1 (ii) Reqd prob = F(0.6) F(0.4) or xxx d)1(20
6.0
4.0
3 M1 = 5454 4.044.056.046.05 + A1
= 0.25 (cao) A1
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(iii)The upper quartile satisfies
4345 54 = qq M1A1
032016 45 =+ qq AG
8 (a) P(X = 3) = 207.0!3
75.3e3
75.3 = M1A1 (b)(i) P(X 5) = 0.0959 M1A1 (ii) Using the Poisson table backwards, M1 = 3.2 A1
(c)(i) P(No errors on 1 page) = 6.0e M1 Reqd prob = ( )n6.0e = n6.0e A1 (ii) We require 01.0e 6.0 n EITHER
Proceed by trial and error. M1 For n = 7, ...014.0eor ...7.66e 6.00.6 == nn A1 For n = 8, ...0082.0eor ...5.121e 6.06.0 == nn A1
Minimum n = 8 A1 OR Solve 0.6n = ln100 (oe) M1A1
n = ..67.76.0
100ln = A1 Minimum n = 8 A1
GCE M/S Mathematics (Jan 2007)/JD
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Welsh Joint Education Committee 245 Western Avenue Cardiff. CF5 2YX Tel. No. 029 2026 5000 Fax. 029 2057 5994 E-mail: [email protected] website: www.wjec.co.uk/exams.html