| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability distribution table |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question requiring standard calculations: verifying a given probability using the binomial formula, completing a probability table, finding expectation, and applying a normal approximation. All techniques are routine for S1 level with no novel problem-solving required, 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 binomial5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 81 } { 256 }\) | \(\frac { 3 } { 64 }\) | \(\frac { 1 } { 256 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(X=3) =]\ \frac{3}{4} \times \left(\frac{1}{4}\right)^3 \times 4\) | M1 | \(\frac{3}{4} \times \left(\frac{1}{4}\right)^3 \times q\); \(q\) a positive integer (1 may be implied) |
| \(= \frac{3}{64}\) | A1 | AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): 0, 1, 2, 3, 4; \(P(X=x)\): \(\frac{81}{256}\), \(\frac{27}{64}\), \(\frac{27}{128}\), \(\frac{3}{64}\), \(\frac{1}{256}\) | B1 | Either \(P(1) = \frac{27}{64} = 0.421875\) or \(P(2) = \frac{27}{128} = 0.2109375\) correct to at least 3SF. Condone not in table |
| B1 FT | Both values in table. FT \(P(1) + P(2) = \frac{81}{128} = 0.6328125\) | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([E(X) =]\ \left[0 \times \frac{81}{256}\right] + 1 \times \textit{their}\frac{27}{64} + 2 \times \textit{their}\frac{27}{128} + 3 \times \frac{12}{256} + 4 \times \frac{1}{256}\) | M1 | Correct method from *their* probability distribution table with at least 4 terms, \(0 < \textit{their}\ P(x) < 1\), accept partially evaluated. \(= 0 + \frac{27}{64} + \frac{54}{128} + \frac{36}{256} + \frac{4}{256}\) |
| \(= 1\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mean} = 96 \times \frac{67}{256} = 25.125\) | B1 | \(25.125\), \(25\frac{1}{8}\) and \(18.5493\ldots\) to at least 3SF seen, allow unsimplified (\(4.3068 \leqslant \sigma \leqslant 4.307\) implies correct variance) |
| \(\text{Var} = 96 \times \frac{67}{256} \times \frac{189}{256} = 18.549\) | ||
| \(P(X < 20) = P\left(Z < \frac{19.5 - 25.125}{\sqrt{18.549}}\right)\) | M1 | Substituting their \(\mu\) and \(\sigma\) into \(\pm\) standardisation formula (any number for 19.5). Condone \(\sigma^2\) and \(\sqrt{\sigma}\) |
| M1 | Using continuity correction 19.5 or 20.5 in their standardisation formula. Note: \(\frac{\pm 5.625}{\sqrt{18.549}}\) seen gains M2 BOD | |
| \([= P(Z < -1.306) = 1 - \Phi(1.306) =] \ 1 - 0.9042 =\) | M1 | Appropriate area \(\Phi\), from final process. Must be a probability |
| \(0.0958\) | A1 | \(0.0957 \leqslant p \leqslant 0.0958\). SC B1 for \(0.0957 \leqslant p \leqslant 0.0958\) if B1M0M0M1 scored |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(X=3) =]\ \frac{3}{4} \times \left(\frac{1}{4}\right)^3 \times 4$ | M1 | $\frac{3}{4} \times \left(\frac{1}{4}\right)^3 \times q$; $q$ a positive integer (1 may be implied) |
| $= \frac{3}{64}$ | A1 | AG |
| | **2** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: 0, 1, 2, 3, 4; $P(X=x)$: $\frac{81}{256}$, $\frac{27}{64}$, $\frac{27}{128}$, $\frac{3}{64}$, $\frac{1}{256}$ | B1 | Either $P(1) = \frac{27}{64} = 0.421875$ or $P(2) = \frac{27}{128} = 0.2109375$ correct to at least 3SF. Condone not in table |
| | B1 FT | Both values in table. FT $P(1) + P(2) = \frac{81}{128} = 0.6328125$ |
| | **2** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[E(X) =]\ \left[0 \times \frac{81}{256}\right] + 1 \times \textit{their}\frac{27}{64} + 2 \times \textit{their}\frac{27}{128} + 3 \times \frac{12}{256} + 4 \times \frac{1}{256}$ | M1 | Correct method from *their* probability distribution table with at least 4 terms, $0 < \textit{their}\ P(x) < 1$, accept partially evaluated. $= 0 + \frac{27}{64} + \frac{54}{128} + \frac{36}{256} + \frac{4}{256}$ |
| $= 1$ | A1 | |
| | **2** | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = 96 \times \frac{67}{256} = 25.125$ | B1 | $25.125$, $25\frac{1}{8}$ and $18.5493\ldots$ to at least 3SF seen, allow unsimplified ($4.3068 \leqslant \sigma \leqslant 4.307$ implies correct variance) |
| $\text{Var} = 96 \times \frac{67}{256} \times \frac{189}{256} = 18.549$ | | |
| $P(X < 20) = P\left(Z < \frac{19.5 - 25.125}{\sqrt{18.549}}\right)$ | M1 | Substituting their $\mu$ and $\sigma$ into $\pm$ standardisation formula (any number for 19.5). Condone $\sigma^2$ and $\sqrt{\sigma}$ |
| | M1 | Using continuity correction 19.5 or 20.5 in their standardisation formula. Note: $\frac{\pm 5.625}{\sqrt{18.549}}$ seen gains M2 BOD |
| $[= P(Z < -1.306) = 1 - \Phi(1.306) =] \ 1 - 0.9042 =$ | M1 | Appropriate area $\Phi$, from final process. Must be a probability |
| $0.0958$ | A1 | $0.0957 \leqslant p \leqslant 0.0958$. **SC B1** for $0.0957 \leqslant p \leqslant 0.0958$ if B1M0M0M1 scored |
---6 Eli has four fair 4 -sided dice with sides labelled $1,2,3,4$. He throws all four dice at the same time. The random variable $X$ denotes the number of 2s obtained.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 3 ) = \frac { 3 } { 64 }$.
\item Complete the following probability distribution table for $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 81 } { 256 }$ & & & $\frac { 3 } { 64 }$ & $\frac { 1 } { 256 }$ \\
\hline
\end{tabular}
\end{center}
\item Find $\mathrm { E } ( X )$.\\
Eli throws the four dice at the same time on 96 occasions.
\item Use an approximation to find the probability that he obtains at least two 2 s on fewer than 20 of these occasions.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q6 [11]}}