| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Multiple independent binomial calculations |
| Difficulty | Moderate -0.3 This question involves standard binomial distribution calculations with straightforward application of formulas. Part (a) is direct cumulative probability, part (b) is geometric distribution (a standard extension), and part (c) is routine normal approximation. All parts require only recall of standard techniques with no novel problem-solving, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - P(10, 11, 12) = 1 - [^{12}C_{10}\ 0.72^{10}\ 0.28^2 + ^{12}C_{11}\ 0.72^{11}\ 0.28^1 + 0.72^{12}]\) | M1 | |
| \(1 - (0.19372 + 0.09057 + 0.01941)\) | A1 | |
| \(0.696\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.28^3 \times 0.72 = 0.0158\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= 100 \times 0.72 = 72\); Var \(= 100 \times 0.72 \times 0.28 = 20.16\) | M1 | |
| \(P(\text{less than } 64) = P\left(z < \frac{63.5 - 72}{\sqrt{20.16}}\right)\) | M1 | M1 for substituting *their* \(\mu\) and \(\sigma\) into \(\pm\)standardisation formula with a numerical value for '63.5' |
| Using either 63.5 or 64.5 within a \(\pm\)standardisation formula | M1 | |
| Appropriate area \(\Phi\), from standardisation formula \(P(z <\ldots)\) in final solution \(= P(z < -1.893)\) | M1 | |
| \(0.0292\) | A1 |
# Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - P(10, 11, 12) = 1 - [^{12}C_{10}\ 0.72^{10}\ 0.28^2 + ^{12}C_{11}\ 0.72^{11}\ 0.28^1 + 0.72^{12}]$ | M1 | |
| $1 - (0.19372 + 0.09057 + 0.01941)$ | A1 | |
| $0.696$ | A1 | |
---
# Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.28^3 \times 0.72 = 0.0158$ | B1 | |
---
# Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 100 \times 0.72 = 72$; Var $= 100 \times 0.72 \times 0.28 = 20.16$ | M1 | |
| $P(\text{less than } 64) = P\left(z < \frac{63.5 - 72}{\sqrt{20.16}}\right)$ | M1 | M1 for substituting *their* $\mu$ and $\sigma$ into $\pm$standardisation formula with a numerical value for '63.5' |
| Using either 63.5 or 64.5 within a $\pm$standardisation formula | M1 | |
| Appropriate area $\Phi$, from standardisation formula $P(z <\ldots)$ in final solution $= P(z < -1.893)$ | M1 | |
| $0.0292$ | A1 | |7 On any given day, the probability that Moena messages her friend Pasha is 0.72 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that for a random sample of 12 days Moena messages Pasha on no more than 9 days.
\item Moena messages Pasha on 1 January. Find the probability that the next day on which she messages Pasha is 5 January.
\item Use an approximation to find the probability that in any period of 100 days Moena messages Pasha on fewer than 64 days.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q7 [9]}}