| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Combined probability with other distributions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question combining standard binomial and normal distribution calculations. Part (a) requires basic binomial probability with clear parameters (n=10, p=0.1), part (b) is routine normal distribution with z-score calculation, and part (c) involves inverse normal lookup. All parts are textbook exercises requiring only direct application of formulas with no problem-solving insight or complex reasoning. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(0,1,2)] = {}^{10}C_0\ 0.1^0\ 0.9^{10} + {}^{10}C_1\ 0.1^1\ 0.9^9 + {}^{10}C_2\ 0.1^2\ 0.9^8\) | M1 | One term \({}^{10}C_x\ p^x(1-p)^{10-x}\), \(0 < p < 1, x \neq 0\) |
| \(= 0.348678 + 0.38742 + 0.19371\) | A1 | Correct expression, accept unsimplified |
| \(0.930\) | B1 | \(0.9298 \leqslant p \leqslant 0.9303\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X > 1.11)] = P\left(Z > \frac{1.11 - 1.04}{0.06}\right) = P(Z > 1.167)\) | M1 | 1.11, 1.04 and 0.06 substituted into \(\pm\)standardisation formula, no continuity correction, not \(0.06^2\) or \(\sqrt{0.06}\) |
| \(= 1 - 0.8784\) | M1 | \(1 - \textit{their}\ 0.8784\) as final answer, must be probability (expect final ans \(< 0.5\)) |
| \(0.122\) | A1 | \(0.1216 \leqslant p \leqslant 0.122\); SC M0 M1 B1 for 0.122 with no standardisation formula |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[P(X < w) = P\left(Z < \frac{w - 1.04}{0.06}\right) = 0.81\right]\) | B1 | \(0.8775 < z \leqslant 0.878\) or \(-0.878 \leqslant z < -0.8775\) seen |
| \(\frac{w - 1.04}{0.06} = 0.878\) | M1 | 1.04 and 0.06 substituted in \(\pm\)standardisation formula, no continuity correction, not \(\sigma^2\), \(\sqrt{\sigma}\), equated to a \(z\)-value |
| \(w = 1.09\) | A1 | \(1.09 \leqslant w \leqslant 1.093\) |
# Question 5(a):
$[P(0,1,2)] = {}^{10}C_0\ 0.1^0\ 0.9^{10} + {}^{10}C_1\ 0.1^1\ 0.9^9 + {}^{10}C_2\ 0.1^2\ 0.9^8$ | M1 | One term ${}^{10}C_x\ p^x(1-p)^{10-x}$, $0 < p < 1, x \neq 0$
$= 0.348678 + 0.38742 + 0.19371$ | A1 | Correct expression, accept unsimplified
$0.930$ | B1 | $0.9298 \leqslant p \leqslant 0.9303$
---
# Question 5(b):
$[P(X > 1.11)] = P\left(Z > \frac{1.11 - 1.04}{0.06}\right) = P(Z > 1.167)$ | M1 | 1.11, 1.04 and 0.06 substituted into $\pm$standardisation formula, no continuity correction, not $0.06^2$ or $\sqrt{0.06}$
$= 1 - 0.8784$ | M1 | $1 - \textit{their}\ 0.8784$ as final answer, must be probability (expect final ans $< 0.5$)
$0.122$ | A1 | $0.1216 \leqslant p \leqslant 0.122$; **SC M0 M1 B1** for 0.122 with no standardisation formula
---
# Question 5(c):
$\left[P(X < w) = P\left(Z < \frac{w - 1.04}{0.06}\right) = 0.81\right]$ | B1 | $0.8775 < z \leqslant 0.878$ or $-0.878 \leqslant z < -0.8775$ seen
$\frac{w - 1.04}{0.06} = 0.878$ | M1 | 1.04 and 0.06 substituted in $\pm$standardisation formula, no continuity correction, not $\sigma^2$, $\sqrt{\sigma}$, equated to a $z$-value
$w = 1.09$ | A1 | $1.09 \leqslant w \leqslant 1.093$
---5 Company $A$ produces bags of sugar. An inspector finds that on average $10 \%$ of the bags are underweight.
10 of the bags are chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that fewer than 3 of these bags are underweight.\\
The weights of the bags of sugar produced by company $B$ are normally distributed with mean 1.04 kg and standard deviation 0.06 kg .
\item Find the probability that a randomly chosen bag produced by company $B$ weighs more than 1.11 kg .\\
$81 \%$ of the bags of sugar produced by company $B$ weigh less than $w \mathrm {~kg}$.
\item Find the value of $w$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q5 [9]}}