| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 Part (a) is a straightforward binomial probability calculation with small n=7, requiring basic calculator work. Part (b) involves a standard normal approximation to binomial with continuity correction—a routine S1 technique with no conceptual challenges. Both parts are slightly easier than average A-level questions due to their mechanical nature and clear setup. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.65^7 + {^7C_1}\, 0.65^6\, 0.35^1 + {^7C_2}\, 0.65^5\, 0.35^2\) | M1 | Binomial term of form \(^7C_x\, p^x(1-p)^{7-x}\), \(0 < p < 1\), any \(p\), \(x \neq 0, 7\) |
| \(0.049022 + 0.184776 + 0.29848\) | A1 | Correct unsimplified answer |
| 0.532 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= 142 \times 0.35 = 49.7\) | B1 | Correct unsimplified \(np\) and \(npq\) (condone \(\sigma = 5.684\) evaluated) |
| Variance \(= 142 \times 0.35 \times 0.65 = 32.305\) | ||
| \(P(X > 40) = P\!\left(z > \dfrac{40.5 - 49.7}{\sqrt{32.305}}\right)\) | M1 | Substituting their \(\mu\) and \(\sigma\) (no \(\sqrt{\sigma}\) or \(\sigma^2\)) into \(\pm\)standardisation formula with numerical value for 40.5 |
| \(P(z > -1.619)\) | M1 | Using either 40.5 or 39.5 within a \(\pm\)standardisation formula |
| M1 | Appropriate area \(\Phi\), from standardisation formula \(P(z > \ldots)\) in final solution, must be probability | |
| 0.947 | A1 | Correct final answer |
## Question 4:
### Part 4(a):
$0.65^7 + {^7C_1}\, 0.65^6\, 0.35^1 + {^7C_2}\, 0.65^5\, 0.35^2$ | M1 | Binomial term of form $^7C_x\, p^x(1-p)^{7-x}$, $0 < p < 1$, any $p$, $x \neq 0, 7$
$0.049022 + 0.184776 + 0.29848$ | A1 | Correct unsimplified answer
0.532 | A1 |
### Part 4(b):
Mean $= 142 \times 0.35 = 49.7$ | B1 | Correct unsimplified $np$ and $npq$ (condone $\sigma = 5.684$ evaluated)
Variance $= 142 \times 0.35 \times 0.65 = 32.305$ | |
$P(X > 40) = P\!\left(z > \dfrac{40.5 - 49.7}{\sqrt{32.305}}\right)$ | M1 | Substituting their $\mu$ and $\sigma$ (no $\sqrt{\sigma}$ or $\sigma^2$) into $\pm$standardisation formula with numerical value for 40.5
$P(z > -1.619)$ | M1 | Using either 40.5 or 39.5 within a $\pm$standardisation formula
| M1 | Appropriate area $\Phi$, from standardisation formula $P(z > \ldots)$ in final solution, must be probability
0.947 | A1 | Correct final answer
---4 The 1300 train from Jahor to Keman runs every day. The probability that the train arrives late in Keman is 0.35 .
\begin{enumerate}[label=(\alph*)]
\item For a random sample of 7 days, find the probability that the train arrives late on fewer than 3 days.\\
A random sample of 142 days is taken.
\item Use an approximation to find the probability that the train arrives late on more than 40 days.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q4 [8]}}