| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard z-score calculations and inverse normal lookups. All three parts are routine textbook exercises requiring only direct use of formulas and tables, with no problem-solving insight or multi-step reasoning beyond basic arithmetic. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 11.3) = P\!\left(z > \frac{11.3 - 10.1}{1.3}\right) = P(z > 0.9231)\) | M1 | Using \(\pm\) standardisation formula, no \(\sqrt{\sigma}\) or \(\sigma^2\), continuity correction |
| \(1 - 0.822\) | M1 | Appropriate area \(\Phi\), from standardisation formula \(P(z>\ldots)\) in final solution |
| \(0.178\) | A1 | \(0.1779\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = -0.674\) | B1 | \(\pm 0.674\) seen (critical value) |
| \(\frac{t - 10.1}{1.3} = -0.674\) | M1 | An equation using \(\pm\) standardisation formula with a \(z\)-value, condone \(\sqrt{\sigma}\) or \(\sigma^2\), continuity correction |
| \(t = 9.22\) | A1 | AWRT. Only dependent on M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(8.9 < X < 11.3) = 1 - 2 \times \text{their } \mathbf{3(a)}\) \(\equiv 2(1 - \text{their } \mathbf{3(a)}) - 1\) \(\equiv 2(0.5 - \text{their } \mathbf{3(a)})\) \(= 0.644\) | B1 FT | FT from their \(\mathbf{3(a)} < 0.5\) or correct, accept unevaluated probability OE |
| Number of days \(= 90 \times 0.644 = 57.96\) | M1 | \(90 \times \text{their } p\) seen, \(0 < p < 1\) |
| So 57 (days) | A1 FT | Accept 57 or 58, not 57.0 or 58.0, no approximation/rounding stated; FT must be an integer value |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\!\left(\frac{8.9-10.1}{1.3} < z < \frac{11.3-10.1}{1.3}\right) = \Phi(0.9231) - (1 - \Phi(0.9231)) = 0.822-(1-0.822) = 0.644\) | B1 | Accept unevaluated probability |
| Number of days \(= 90 \times 0.644 = 57.96\) | M1 | \(90 \times \text{their } p\) seen, \(0 < p < 1\) |
| So 57 (days) | A1 FT | Accept 57 or 58; FT must be an integer value |
## Question 3(a):
$P(X > 11.3) = P\!\left(z > \frac{11.3 - 10.1}{1.3}\right) = P(z > 0.9231)$ | M1 | Using $\pm$ standardisation formula, no $\sqrt{\sigma}$ or $\sigma^2$, continuity correction
$1 - 0.822$ | M1 | Appropriate area $\Phi$, from standardisation formula $P(z>\ldots)$ in final solution
$0.178$ | A1 | $0.1779\ldots$
---
## Question 3(b):
$z = -0.674$ | B1 | $\pm 0.674$ seen (critical value)
$\frac{t - 10.1}{1.3} = -0.674$ | M1 | An equation using $\pm$ standardisation formula with a $z$-value, condone $\sqrt{\sigma}$ or $\sigma^2$, continuity correction
$t = 9.22$ | A1 | AWRT. Only dependent on M1
---
## Question 3(c):
$P(8.9 < X < 11.3) = 1 - 2 \times \text{their } \mathbf{3(a)}$ $\equiv 2(1 - \text{their } \mathbf{3(a)}) - 1$ $\equiv 2(0.5 - \text{their } \mathbf{3(a)})$ $= 0.644$ | B1 FT | FT from their $\mathbf{3(a)} < 0.5$ or correct, accept unevaluated probability OE
Number of days $= 90 \times 0.644 = 57.96$ | M1 | $90 \times \text{their } p$ seen, $0 < p < 1$
So 57 (days) | A1 FT | Accept 57 or 58, not 57.0 or 58.0, no approximation/rounding stated; FT must be an integer value
**Alternative method:**
$P\!\left(\frac{8.9-10.1}{1.3} < z < \frac{11.3-10.1}{1.3}\right) = \Phi(0.9231) - (1 - \Phi(0.9231)) = 0.822-(1-0.822) = 0.644$ | B1 | Accept unevaluated probability
Number of days $= 90 \times 0.644 = 57.96$ | M1 | $90 \times \text{their } p$ seen, $0 < p < 1$
So 57 (days) | A1 FT | Accept 57 or 58; FT must be an integer value
---3 Pia runs 2 km every day and her times in minutes are normally distributed with mean 10.1 and standard deviation 1.3.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that on a randomly chosen day Pia takes longer than 11.3 minutes to run 2 km .
\item On $75 \%$ of days, Pia takes longer than $t$ minutes to run 2 km . Find the value of $t$.
\item On how many days in a period of 90 days would you expect Pia to take between 8.9 and 11.3 minutes to run 2 km ?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q3 [9]}}