Use this ZIMSEC A-Level Mathematics Paper 4 November 2003 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC A-Level Mathematics Paper 4 November 2003 Question Paper
This is the ZIMSEC A-Level Mathematics marking scheme for Paper 4 from November 2003. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
MATHS 4 NOV 2003 MARKING SCHEME
States $\mathrm{H}_{0} \mu=15.85$ and $\mathrm{H}_{1} \mu>15.85$ | B1
Uses $\pm \frac{(16.03-15.85) \sqrt{ } 1225}{2.87}$ | M1
Obtains $\mathrm{Z}=12.195$ | A1
States $\mathrm{Z} \alpha= \pm 2.05$ | B1
Clear comparison of attempted Z value with $\mathrm{Z} \alpha= \pm 2.05$ (NB signs consistent)
M1
Correct conclusion from calculated z and $\pm 2.05$ only and $\mathrm{H}_{0} \mathrm{u}=15.85, \mathrm{H}_{1}: \mu$ ) 15.85 (mean is greater than 15.85)
A1
2. (i) Starting Y - Geo (P)
B1
Obtains $E(Y)=\frac{1}{1 / 4}=4$
B1
(ii) Obtains $\mathrm{P}\left(\mathrm{A}\right.$ wins on his $1^{\text {st }}$ turn $)-1 / 4$
B1
Obtains $\mathrm{P}\left(\mathrm{A}\right.$ wins on his $2^{\mathrm{nd}}$ turn $)=(3 / 4)^{2} \times \frac{1}{4}$
B1
Obtains $\mathrm{P}(\mathrm{A}$ wins on the game $)=1 / 4+(3 / 4)^{2} \times 1 / 4+(3 / 4)^{4} \times 1 / 4+$. M1
Identifying sum to infinity $\mathrm{S}_{\infty}$ and using it correctly M1
Obtains $4 / 7[(0,57)$
A1
3. (i) States $\mathrm{P}(\mathrm{x}<\mathrm{b})=0.975$
B1
Uses $\Phi(\underline{\mathrm{b}-\mu}) \quad=\quad 0.975$
M1
( c )
Considers $\frac{\mathrm{b}-\mu}{\mathrm{a}}=1.96$
B1
Obtains $b=\mu+1.96 \sigma$
A1