Use this ZIMSEC Secondary Statistics Paper 2 2019 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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This is the ZIMSEC Statistics marking scheme for Paper 2 from 2019. It shows how marks are awarded, so you can mark your own answers.
2
(a) $\quad \int_{0}^{3} \frac{x}{12} d x+k \int_{3}^{8}(x-8) d x=1$
$\frac{1}{12}\left[\frac{x^{2}}{2}\right]_{0}^{3}+k\left[\frac{x^{2}}{2}-8 x\right]_{3}^{8}=1 \quad$ (Attempt to integrate and equating to 1$) \rightarrow \mathrm{M} 1$
\[
\begin{array}{l}
\frac{3^{2}}{24}+k\left[\frac{8^{2}}{2}-8(8)-\frac{3^{2}}{2}+8(3)\right]=1
k\left(\frac{-25}{2}\right)=\frac{5}{8}
\end{array}
\]
(b) $\quad E(X)=\int_{0}^{3} \frac{x^{2}}{12} d x+\frac{-1}{20} \int_{3}^{8}\left(x^{2}-8 x\right) d x$
\[
\begin{array}{l}
\left.=\frac{1}{12}\left(\frac{x^{3}}{3}\right)\right]_{0}^{3}-\frac{1}{20}\left[\frac{x^{3}}{3}-\frac{8 x^{2}}{2}\right]_{3}^{8} \text { attempto to } \int \operatorname{lis}(x f(x)) \rightarrow M 1
=\frac{1}{36}\left[3^{3}-0^{3}\right]-\frac{1}{20}\left[\frac{8^{3}}{3}-\frac{8(8)^{2}}{2}-\frac{3^{3}}{3}+\frac{8(3)^{2}}{2}\right]
\end{array}
\]
\[
\begin{aligned}
\frac{11}{3} \text { or } \frac{3 \frac{2}{3}}{(c)} | \left.=\int_{0}^{3} \frac{x}{12} d x=\frac{x^{2}}{24}\right]_{0}^{3} | =\frac{3^{2}}{24}-\frac{0^{2}}{24} | =\frac{3}{8}
\end{aligned}
\]
A1
Median $=\frac{-1}{20} \int_{m}^{8}(x-8) d x=0.5$ affempt to lut with correct limits
\[
\begin{array}{l}
\frac{-1}{20}\left[\frac{8^{2}}{2}-8(8)-\frac{m^{2}}{2}+8 m\right]
\frac{m^{2}}{2}-8 m+22=0
m^{2}-16 m+2 \frac{1}{2}=0
m=\frac{-(-16) \pm \sqrt{16^{2}-4(1)(44)}}{2(1)}
\end{array} \quad \frac{\theta}{3} \frac{x}{12} d e+k \int_{3}^{m}(x-0) d x=0,5
\]
atempt to sulvefor $\mathrm{m} \Rightarrow \mathrm{MI}$
\[
=3.528
\]
Al
|7|
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