| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.8 This is a straightforward normal distribution question requiring only standard table lookups and z-score calculations. Part (i) involves calculating P(X < 132) using z = (132-140)/12, and part (ii) requires reverse lookup to find k from P(X > k) = 0.675. Both are routine S1 procedures with no problem-solving or conceptual challenges beyond basic normal distribution mechanics. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 132) = P\!\left(Z < \frac{132-140}{12}\right) = P(Z < -0.6667)\) | M1 | Using \(\pm\) standardisation formula, no continuity correction, not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(= 1 - 0.7477\) | M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z<\ldots)\) in final solution |
| \(= 0.252\) awrt | A1 | Condone linear interpolation \(= 0.25243\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{time}>k) = 0.675\), \(z = -0.454\) | B1 | \(\pm 0.454\) seen |
| \(\frac{k-140}{12} = -0.454\) | M1 | An equation using the standardisation formula with a \(z\)-value (not \(1-z\)), condone \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(k = 135, 134.6, 134.55\) | A1 | B0M1A1 max from \(-0.45\) |
## Question 3:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 132) = P\!\left(Z < \frac{132-140}{12}\right) = P(Z < -0.6667)$ | M1 | Using $\pm$ standardisation formula, no continuity correction, not $\sigma^2$ or $\sqrt{\sigma}$ |
| $= 1 - 0.7477$ | M1 | Appropriate area $\Phi$ from standardisation formula $P(z<\ldots)$ in final solution |
| $= 0.252$ awrt | A1 | Condone linear interpolation $= 0.25243$ |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{time}>k) = 0.675$, $z = -0.454$ | B1 | $\pm 0.454$ seen |
| $\frac{k-140}{12} = -0.454$ | M1 | An equation using the standardisation formula with a $z$-value (not $1-z$), condone $\sigma^2$ or $\sqrt{\sigma}$ |
| $k = 135, 134.6, 134.55$ | A1 | B0M1A1 max from $-0.45$ |
---3 The times taken, in minutes, for trains to travel between Alphaton and Beeton are normally distributed with mean 140 and standard deviation 12.\\
(i) Find the probability that a randomly chosen train will take less than 132 minutes to travel between Alphaton and Beeton.\\
(ii) The probability that a randomly chosen train takes more than $k$ minutes to travel between Alphaton and Beeton is 0.675 . Find the value of $k$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q3 [6]}}