Moderate -0.5 This is a standard textbook application of the normal approximation to the binomial distribution with continuity correction. It requires identifying n=120, p=0.4, calculating mean and variance, applying continuity correction (35.5 to 54.5), and using normal tables—all routine procedures for S1 level with no problem-solving insight needed. Slightly easier than average due to its straightforward nature.
2 Anil is a candidate in an election. He received \(40 \%\) of the votes. A random sample of 120 voters is chosen.
Use an approximation to find the probability that, of the 120 voters, between 36 and 54 inclusive voted for Anil.
Mean \(= 120 \times 0.4 = 48\), Var \(= 120 \times 0.4 \times 0.6 = 28.8\)
B1
48 and \(28\frac{4}{5}\), 28.8 seen, allow unsimplified. \((5.366 \leq \sigma \leq 5.367\) or \(\frac{12\sqrt{5}}{5}\) implies correct variance)
\(P(36 \leq X \leq 54) = P\left(\frac{35.5-48}{\sqrt{28.8}} < Z < \frac{54.5-48}{\sqrt{28.8}}\right)\)
M1
Substituting their \(\mu\) and \(\sigma\) into \(\pm\)standardisation formula (any number for 35.5 or 54.5), condone \(\sigma^2\) and \(\sqrt{\sigma}\)
M1
Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula. Note: \(\frac{\pm12.5}{\sqrt{28.8}}\) or \(\frac{\pm6.5}{\sqrt{28.8}}\) seen gains M2 BOD
Appropriate area \(\Phi\), from final process. Must be a probability. Expect final answer \(> 0.5\). Note: correct final answer implies this M1
\(= 0.877\)
A1
\(0.877 \leq p \leq 0.8772\)
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 120 \times 0.4 = 48$, Var $= 120 \times 0.4 \times 0.6 = 28.8$ | B1 | 48 and $28\frac{4}{5}$, 28.8 seen, allow unsimplified. $(5.366 \leq \sigma \leq 5.367$ or $\frac{12\sqrt{5}}{5}$ implies correct variance) |
| $P(36 \leq X \leq 54) = P\left(\frac{35.5-48}{\sqrt{28.8}} < Z < \frac{54.5-48}{\sqrt{28.8}}\right)$ | M1 | Substituting their $\mu$ and $\sigma$ into $\pm$standardisation formula (any number for 35.5 or 54.5), condone $\sigma^2$ and $\sqrt{\sigma}$ |
| | M1 | Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula. Note: $\frac{\pm12.5}{\sqrt{28.8}}$ or $\frac{\pm6.5}{\sqrt{28.8}}$ seen gains M2 BOD |
| $[= P(-2.3292 < Z < 1.211) =] \ 0.8871 + 0.9900 - 1$ | M1 | Appropriate area $\Phi$, from final process. Must be a probability. Expect final answer $> 0.5$. Note: correct final answer implies this M1 |
| $= 0.877$ | A1 | $0.877 \leq p \leq 0.8772$ |
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2 Anil is a candidate in an election. He received $40 \%$ of the votes. A random sample of 120 voters is chosen.
Use an approximation to find the probability that, of the 120 voters, between 36 and 54 inclusive voted for Anil.\\
\hfill \mbox{\textit{CAIE S1 2023 Q2 [5]}}