CAIE S1 2023 November — Question 3 5 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2023
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeSingle probability inequality
DifficultyModerate -0.8 This is a straightforward application of the normal approximation to the binomial distribution with continuity correction. Students need to identify n=200, p=0.15, calculate mean and variance, apply continuity correction (P(X>40) = P(X≥41) → P(Y>40.5)), and use standard normal tables. It's a routine textbook exercise requiring only procedural knowledge with no problem-solving insight needed.
Spec2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

3 A factory produces a certain type of electrical component. It is known that \(15 \%\) of the components produced are faulty. A random sample of 200 components is chosen. Use an approximation to find the probability that more than 40 of these components are faulty.

Question 3:
AnswerMarks Guidance
AnswerMark Guidance
\([\text{Mean} = 200\times0.15 =]\ 30\); \([\text{Var} = 200\times0.15\times0.85 =]\ 25.5\)B1 30 and 25.5, \(25\frac{1}{2}, \frac{51}{2}\) seen, allow unsimplified. May be seen in standardisation formula. \([\sigma =]\ 5.049 \leqslant \sigma \leqslant 5.05[0]\), \(\frac{\sqrt{102}}{2}\) implies correct variance. Correct notation required.
\([P(X>40) =]\ P\!\left(Z > \dfrac{40.5-30}{\sqrt{25.5}}\right)\)M1 Substituting *their* mean and *their* positive 5.04975 into \(\pm\)standardisation formula (any number for 40.5), not *their* \(\sigma^2\) or \(\sqrt{\textit{their}\ \sigma}\)
M1Using continuity correction 39.5 or 40.5 in *their* standardisation formula
\([1 - \Phi(2.079)]\ 1 - 0.9812\)M1 Appropriate area \(\Phi\), from final process, must be a probability
\(= 0.0188\)A1 \(0.01875 < p \leqslant 0.0188\)
5
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Mean} = 200\times0.15 =]\ 30$; $[\text{Var} = 200\times0.15\times0.85 =]\ 25.5$ | B1 | 30 and 25.5, $25\frac{1}{2}, \frac{51}{2}$ seen, allow unsimplified. May be seen in standardisation formula. $[\sigma =]\ 5.049 \leqslant \sigma \leqslant 5.05[0]$, $\frac{\sqrt{102}}{2}$ implies correct variance. Correct notation required. |
| $[P(X>40) =]\ P\!\left(Z > \dfrac{40.5-30}{\sqrt{25.5}}\right)$ | M1 | Substituting *their* mean and *their* positive 5.04975 into $\pm$standardisation formula (any number for 40.5), not *their* $\sigma^2$ or $\sqrt{\textit{their}\ \sigma}$ |
| | M1 | Using continuity correction 39.5 or 40.5 in *their* standardisation formula |
| $[1 - \Phi(2.079)]\ 1 - 0.9812$ | M1 | Appropriate area $\Phi$, from final process, must be a probability |
| $= 0.0188$ | A1 | $0.01875 < p \leqslant 0.0188$ |
| | **5** | |

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3 A factory produces a certain type of electrical component. It is known that $15 \%$ of the components produced are faulty. A random sample of 200 components is chosen.

Use an approximation to find the probability that more than 40 of these components are faulty.\\

\hfill \mbox{\textit{CAIE S1 2023 Q3 [5]}}