| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard techniques: z-score calculations, inverse normal lookups, and using symmetry properties. Part (b) adds mild algebraic manipulation (expressing σ in terms of μ) but all parts are routine S1-level applications with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 4) = P\!\left(Z < \frac{4-3.24}{0.96}\right)\) | M1 | \(\pm\)Standardisation formula, no cc, no sq rt, no square |
| \(= P(Z < 0.7917) = 0.7858\) | A1 | \(0.7855 < p \leqslant 0.7858\) or \(p = 0.786\) cao (implies M1A1 awarded), may be seen used in calculation |
| *their* \(0.7858 \times 365 = 286\) (or 287) | B1ft | *Their* probability \(\times 365\) provided 4sf probability seen. FT answer rounded or truncated to nearest integer. No approximation notation used. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < k) = P\!\left(Z < \frac{k-3.24}{0.96}\right) = 0.8\) | B1 | \((z=) \pm 0.842\) seen |
| \(\frac{k-3.24}{0.96} = 0.842\) | M1 | \(z = \pm\frac{k-3.24}{0.96}\), allow cc, sq rt or square equated to a \(z\)-value (0.7881, 0.2119, 0.158, 0.8, 0.2 etc. are not acceptable) |
| \(k = 4.05\) | A1 | Correct final answer, www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(-1.5 < Z < 1.5) =\) | M1 | \(\Phi(z=1.5)\) or \(\Phi(z=-1.5)\) seen used or \(p = 0.9332\) seen |
| \(\Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1 = 2 \times 0.9332 - 1\) oe | M1 | Correct final area expression using *their* probabilities |
| \(= 0.866\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(Y>0) = P\!\left(Z > \frac{0-\mu}{\sigma}\right) \equiv P\!\left(Z > \frac{0-\mu}{3\mu/4}\right)\) or \(P\!\left(Z > \frac{0-\left(\frac{4\sigma}{3}\right)}{\sigma}\right)\) | M1 | \(\pm\)Standardisation attempt in terms of one variable, no sq rt or square, condone \(\pm 0.5\) as cc |
| \(= P(Z > -4/3)\) | A1 | Correct unsimplified standardisation, no variables |
| \(= 0.909\) | A1 | Correct final answer |
## Question 7(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 4) = P\!\left(Z < \frac{4-3.24}{0.96}\right)$ | M1 | $\pm$Standardisation formula, no cc, no sq rt, no square |
| $= P(Z < 0.7917) = 0.7858$ | A1 | $0.7855 < p \leqslant 0.7858$ or $p = 0.786$ cao (implies M1A1 awarded), may be seen used in calculation |
| *their* $0.7858 \times 365 = 286$ (or 287) | B1ft | *Their* probability $\times 365$ provided 4sf probability **seen**. FT answer rounded or truncated to nearest integer. No approximation notation used. |
**Total: 3 marks**
---
## Question 7(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < k) = P\!\left(Z < \frac{k-3.24}{0.96}\right) = 0.8$ | B1 | $(z=) \pm 0.842$ seen |
| $\frac{k-3.24}{0.96} = 0.842$ | M1 | $z = \pm\frac{k-3.24}{0.96}$, allow cc, sq rt or square equated to a $z$-value (0.7881, 0.2119, 0.158, 0.8, 0.2 etc. are not acceptable) |
| $k = 4.05$ | A1 | Correct final answer, www |
**Total: 3 marks**
---
## Question 7(a)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(-1.5 < Z < 1.5) =$ | M1 | $\Phi(z=1.5)$ or $\Phi(z=-1.5)$ seen used or $p = 0.9332$ seen |
| $\Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1 = 2 \times 0.9332 - 1$ oe | M1 | Correct final area expression using *their* probabilities |
| $= 0.866$ | A1 | Correct final answer |
**Total: 3 marks**
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(Y>0) = P\!\left(Z > \frac{0-\mu}{\sigma}\right) \equiv P\!\left(Z > \frac{0-\mu}{3\mu/4}\right)$ or $P\!\left(Z > \frac{0-\left(\frac{4\sigma}{3}\right)}{\sigma}\right)$ | M1 | $\pm$Standardisation attempt in terms of one variable, no sq rt or square, condone $\pm 0.5$ as cc |
| $= P(Z > -4/3)$ | A1 | Correct unsimplified standardisation, no variables |
| $= 0.909$ | A1 | Correct final answer |
**Total: 3 marks**
---7
\begin{enumerate}[label=(\alph*)]
\item The time, $X$ hours, for which students use a games machine in any given day has a normal distribution with mean 3.24 hours and standard deviation 0.96 hours.
\begin{enumerate}[label=(\roman*)]
\item On how many days of the year ( 365 days) would you expect a randomly chosen student to use a games machine for less than 4 hours?
\item Find the value of $k$ such that $\mathrm { P } ( X > k ) = 0.2$.
\item Find the probability that the number of hours for which a randomly chosen student uses a games machine in a day is within 1.5 standard deviations of the mean.
\end{enumerate}\item The variable $Y$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, where $4 \sigma = 3 \mu$ and $\mu \neq 0$. Find the probability that a randomly chosen value of $Y$ is positive.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2018 Q7 [12]}}