| 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 procedures: (a) requires a single z-score calculation and table lookup, (b) involves inverse normal (finding a value given a probability), and (c) combines two probabilities then multiplies by 365. All three parts are routine textbook exercises requiring only direct application of formulas 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X > 4.2) = P\!\left(z > \dfrac{4.2 - 3.5}{0.9}\right) = P(z > 0.7778)\) | M1 | Using \(\pm\) standardisation formula, no \(\sqrt{\sigma}\) or \(\sigma^2\), continuity correction |
| \(1 - 0.7818\) | M1 | Appropriate area \(\Phi\), from standardisation formula \(P(z>\ldots)\) in final solution |
| \(0.218\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = -1.282\) | B1 | \(\pm 1.282\) seen (critical value) |
| \(\dfrac{t - 3.5}{0.9} = -1.282\) | M1 | An equation using \(\pm\) standardisation formula with a \(z\)-value, condone \(\sqrt{\sigma}\), \(\sigma^2\) and continuity correction |
| \(t = 2.35\) | A1 | AWRT, only dependent on M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2.8 < X < 4.2) = 1 - 2 \times \text{their } \mathbf{5(a)} \equiv 2(1 - \text{their } \mathbf{5(a)}) - 1 \equiv 2(0.5 - \text{their } \mathbf{5(a)}) = 0.5636\) | B1 FT | FT from their \(\mathbf{5(a)} < 0.5\) or correct. Accept unevaluated probability OE. Accept \(0.564\) |
| Number of days \(= 365 \times 0.5636 = 205.7\) | M1 | \(365 \times \text{their } p\) |
| So, \(205\) (days) | A1 FT | Accept 205 or 206, not 205.0 or 206.0; no approximation/rounding stated. FT must be an integer value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\!\left(\dfrac{2.8-3.5}{0.9} < z < \dfrac{4.2-3.5}{0.9}\right) = \Phi(0.7778)-(1-\Phi(0.7778)) = 0.7818-(1-0.7818) = 0.5636\) | B1 | \(0.5635 < p \leqslant 0.564\) OE |
| Number of days \(= 365 \times 0.5636 = 205.7\) | M1 | \(365 \times \text{their } p\) |
| So, \(205\) (days) | A1 FT | Accept 205 or 206; FT must be an integer value |
## Question 5:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 4.2) = P\!\left(z > \dfrac{4.2 - 3.5}{0.9}\right) = P(z > 0.7778)$ | M1 | Using $\pm$ standardisation formula, no $\sqrt{\sigma}$ or $\sigma^2$, continuity correction |
| $1 - 0.7818$ | M1 | Appropriate area $\Phi$, from standardisation formula $P(z>\ldots)$ in final solution |
| $0.218$ | A1 | |
**Total: 3 marks**
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = -1.282$ | B1 | $\pm 1.282$ seen (critical value) |
| $\dfrac{t - 3.5}{0.9} = -1.282$ | M1 | An equation using $\pm$ standardisation formula with a $z$-value, condone $\sqrt{\sigma}$, $\sigma^2$ and continuity correction |
| $t = 2.35$ | A1 | AWRT, only dependent on M mark |
**Total: 3 marks**
**Part (c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2.8 < X < 4.2) = 1 - 2 \times \text{their } \mathbf{5(a)} \equiv 2(1 - \text{their } \mathbf{5(a)}) - 1 \equiv 2(0.5 - \text{their } \mathbf{5(a)}) = 0.5636$ | B1 FT | FT from their $\mathbf{5(a)} < 0.5$ or correct. Accept unevaluated probability OE. Accept $0.564$ |
| Number of days $= 365 \times 0.5636 = 205.7$ | M1 | $365 \times \text{their } p$ |
| So, $205$ (days) | A1 FT | Accept 205 or 206, not 205.0 or 206.0; no approximation/rounding stated. FT must be an integer value |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(\dfrac{2.8-3.5}{0.9} < z < \dfrac{4.2-3.5}{0.9}\right) = \Phi(0.7778)-(1-\Phi(0.7778)) = 0.7818-(1-0.7818) = 0.5636$ | B1 | $0.5635 < p \leqslant 0.564$ OE |
| Number of days $= 365 \times 0.5636 = 205.7$ | M1 | $365 \times \text{their } p$ |
| So, $205$ (days) | A1 FT | Accept 205 or 206; FT must be an integer value |
**Total: 3 marks**5 The time in hours that Davin plays on his games machine each day is normally distributed with mean 3.5 and standard deviation 0.9.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that on a randomly chosen day Davin plays on his games machine for more than 4.2 hours.
\item On 90\% of days Davin plays on his games machine for more than $t$ hours. Find the value of $t$.
\item Calculate an estimate for the number of days in a year ( 365 days) on which Davin plays on his games machine for between 2.8 and 4.2 hours.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q5 [9]}}