| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| 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 -0.8 This is a straightforward application of binomial probability and normal approximation with clear parameters. Part (a) is basic probability multiplication, part (b) is standard binomial calculation, and part (c) is a routine normal approximation with continuity correction. All steps are textbook procedures requiring no problem-solving insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([(0.7)^3 =] 0.343\) | B1 | Evaluated WWW |
| Alternative: \([(0.15)^3 + {}^3C_1(0.15)^2(0.55) + {}^3C_2(0.15)(0.55)^2 + (0.55)^3 =] 0.343\) | B1 | Evaluated WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - (0.85^9 + {}^9C_1\, 0.15^1\, 0.85^8 + {}^9C_2\, 0.15^2\, 0.85^7)\) \([1 - (0.231617 + 0.367862 + 0.259667)]\) | M1 | One term: \({}^9C_x\, p^x\,(1-p)^{9-x}\) for \(0 < x < 9\), any \(0 < p < 1\) |
| A1 | Correct expression, accept unsimplified. | |
| \(0.141\) | A1 | \(0.1408 \leqslant \text{ans} \leqslant 0.141\), award at most accurate value. |
| Alternative: \({}^9C_3\, 0.15^3\, 0.85^6 + {}^9C_4\, 0.15^4\, 0.85^5 + {}^9C_5\, 0.15^5\, 0.85^4 + {}^9C_6\, 0.15^6\, 0.85^3 + {}^9C_7\, 0.15^7\, 0.85^2 + {}^9C_8\, 0.15^8\, 0.85 + 0.15^9\) | M1 | One term: \({}^9C_x\, p^x\,(1-p)^{9-x}\) for \(0 < x < 9\), any \(0 < p < 1\) |
| A1 | Correct expression, accept unsimplified. | |
| \(0.141\) | A1 | \(0.1408 \leqslant \text{ans} \leqslant 0.141\), award at most accurate value. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= [60 \times 0.15 =] 9\), Variance \(= [60 \times 0.15 \times 0.85 =] 7.65\) | B1 | Correct mean and variance, allow unsimplified. (\(2.765 \leq \sigma \leq 2.77\) imply correct variance) |
| \([(X \geq 12) =] P\!\left(Z > \frac{11.5 - 9}{\sqrt{7.65}}\right)\) | M1 | Substituting *their* mean and variance into \(\pm\)standardisation formula (any number for 11.5), not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| M1 | Using continuity correction 11.5 or 12.5 in *their* standardisation formula. | |
| \(1 - \Phi(0.9039) = 1 - 0.8169\) | M1 | Appropriate area \(\Phi\), from final process, must be probability. |
| \(0.183\) | A1 | Final AWRT |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[(0.7)^3 =] 0.343$ | B1 | Evaluated WWW |
| **Alternative:** $[(0.15)^3 + {}^3C_1(0.15)^2(0.55) + {}^3C_2(0.15)(0.55)^2 + (0.55)^3 =] 0.343$ | B1 | Evaluated WWW |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - (0.85^9 + {}^9C_1\, 0.15^1\, 0.85^8 + {}^9C_2\, 0.15^2\, 0.85^7)$ $[1 - (0.231617 + 0.367862 + 0.259667)]$ | M1 | One term: ${}^9C_x\, p^x\,(1-p)^{9-x}$ for $0 < x < 9$, any $0 < p < 1$ |
| | A1 | Correct expression, accept unsimplified. |
| $0.141$ | A1 | $0.1408 \leqslant \text{ans} \leqslant 0.141$, award at most accurate value. |
| **Alternative:** ${}^9C_3\, 0.15^3\, 0.85^6 + {}^9C_4\, 0.15^4\, 0.85^5 + {}^9C_5\, 0.15^5\, 0.85^4 + {}^9C_6\, 0.15^6\, 0.85^3 + {}^9C_7\, 0.15^7\, 0.85^2 + {}^9C_8\, 0.15^8\, 0.85 + 0.15^9$ | M1 | One term: ${}^9C_x\, p^x\,(1-p)^{9-x}$ for $0 < x < 9$, any $0 < p < 1$ |
| | A1 | Correct expression, accept unsimplified. |
| $0.141$ | A1 | $0.1408 \leqslant \text{ans} \leqslant 0.141$, award at most accurate value. |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= [60 \times 0.15 =] 9$, Variance $= [60 \times 0.15 \times 0.85 =] 7.65$ | B1 | Correct mean and variance, allow unsimplified. ($2.765 \leq \sigma \leq 2.77$ imply correct variance) |
| $[(X \geq 12) =] P\!\left(Z > \frac{11.5 - 9}{\sqrt{7.65}}\right)$ | M1 | Substituting *their* mean and variance into $\pm$standardisation formula (any number for 11.5), not $\sigma^2$ or $\sqrt{\sigma}$ |
| | M1 | Using continuity correction 11.5 or 12.5 in *their* standardisation formula. |
| $1 - \Phi(0.9039) = 1 - 0.8169$ | M1 | Appropriate area $\Phi$, from final process, must be probability. |
| $0.183$ | A1 | Final AWRT |
---5 Every day Richard takes a flight between Astan and Bejin. On any day, the probability that the flight arrives early is 0.15 , the probability that it arrives on time is 0.55 and the probability that it arrives late is 0.3 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that on each of 3 randomly chosen days, Richard's flight does not arrive late.
\item Find the probability that for 9 randomly chosen days, Richard's flight arrives early at least 3 times.
\item 60 days are chosen at random.
Use an approximation to find the probability that Richard's flight arrives early at least 12 times.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q5 [9]}}