Standard +0.3 This is a straightforward application of the sum of independent normal variables. Students need to recognize that S ~ N(60×3.2, 60×1.2²), then perform a single standardization and table lookup. It's slightly above average difficulty only because it requires knowing the variance scaling rule, but this is a standard S2 technique with no problem-solving insight needed.
2 The random variable \(X\) has the distribution \(\mathrm { N } \left( 3.2,1.2 ^ { 2 } \right)\). The sum of 60 independent observations of \(X\) is denoted by \(S\). Find \(\mathrm { P } ( S > 200 )\).
\(192\) or \(60 \times 3.2\) seen. \(86.4\) or \(60 \times 1.2^2\) or equivalent seen. Standardising with sq rt and no cc. Correct area i.e. \(< 0.5\). Correct answer OR \(200/60\) (B1) \(1.2/\sqrt{60}\) (B1) \(1.2\sqrt{60}\) (M1) etc.
$\sum 60 \text{ obs} \sim N(60 \times 3.2, 60 \times 1.2^2) \sim N(192, 86.4)$
$P(S > 200) = 1 - \Phi\left(\frac{200-192}{\sqrt{86.4}}\right) = 1 - \Phi(0.861) = 0.195$ | B1, B1, M1, M1, A1 | $192$ or $60 \times 3.2$ seen. $86.4$ or $60 \times 1.2^2$ or equivalent seen. Standardising with sq rt and no cc. Correct area i.e. $< 0.5$. Correct answer OR $200/60$ (B1) $1.2/\sqrt{60}$ (B1) $1.2\sqrt{60}$ (M1) etc.
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2 The random variable $X$ has the distribution $\mathrm { N } \left( 3.2,1.2 ^ { 2 } \right)$. The sum of 60 independent observations of $X$ is denoted by $S$. Find $\mathrm { P } ( S > 200 )$.
\hfill \mbox{\textit{CAIE S2 2007 Q2 [5]}}