| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Quadratic relationship μ = kσ² |
| Difficulty | Standard +0.8 Part (a) is a standard reverse normal distribution problem requiring two z-score equations solved simultaneously—routine for S1 but multi-step. Part (b) is more sophisticated: the constraint 3σ = 4μ creates a quadratic relationship requiring algebraic manipulation to express P(X < 3μ) in terms of a standard normal variable, which is less routine and requires insight to recognize the standardization approach. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(z_1 = 2.4\) | B1 | \(\pm 2.4\) seen, accept 2.396 |
| \(z_2 = -0.5\) | B1 | \(\pm 0.5\) seen |
| \(2.4 = \frac{36800 - \mu}{\sigma}\) | M1 | Either standardisation eqn with \(z\) value, not 0.5082, 0.7565, 0.0082, 0.6915, 0.3085, 0.6209, 0.0032 or any other probability |
| \(-0.5 = \frac{31000 - \mu}{\sigma}\) | M1 | Sensible attempt to eliminate \(\mu\) or \(\sigma\) by substitution or subtraction from their 2 equations (\(z\)-value not required), need at least 1 value stated |
| \(\sigma = 2000\), \(\mu = 32000\) | A1 | Both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 3\mu) = P\left(z < \frac{3\mu - \mu}{(4\mu/3)}\right)\) or \(P = \left(z < \frac{(9\sigma/4)-(3\sigma/4)}{\sigma}\right)\) | M1 | Standardise, in terms of one variable, accept \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(P\left(z < \frac{6}{4}\right)\) | M1 | \(\frac{6}{4}\) or \(\frac{6}{4\sigma}\) seen |
| \(= 0.933\) | A1 | Correct final answer |
## Question 4(a):
$z_1 = 2.4$ | B1 | $\pm 2.4$ seen, accept 2.396 |
$z_2 = -0.5$ | B1 | $\pm 0.5$ seen |
$2.4 = \frac{36800 - \mu}{\sigma}$ | M1 | Either standardisation eqn with $z$ value, not 0.5082, 0.7565, 0.0082, 0.6915, 0.3085, 0.6209, 0.0032 or any other probability |
$-0.5 = \frac{31000 - \mu}{\sigma}$ | M1 | Sensible attempt to eliminate $\mu$ or $\sigma$ by substitution or subtraction from their 2 equations ($z$-value not required), need at least 1 value stated |
$\sigma = 2000$, $\mu = 32000$ | A1 | Both correct answers |
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## Question 4(b):
$P(X < 3\mu) = P\left(z < \frac{3\mu - \mu}{(4\mu/3)}\right)$ or $P = \left(z < \frac{(9\sigma/4)-(3\sigma/4)}{\sigma}\right)$ | M1 | Standardise, in terms of one variable, accept $\sigma^2$ or $\sqrt{\sigma}$ |
$P\left(z < \frac{6}{4}\right)$ | M1 | $\frac{6}{4}$ or $\frac{6}{4\sigma}$ seen |
$= 0.933$ | A1 | Correct final answer |
---4
\begin{enumerate}[label=(\alph*)]
\item The distance that car tyres of a certain make can travel before they need to be replaced has a normal distribution. A survey of a large number of these tyres found that the probability of this distance being more than 36800 km is 0.0082 and the probability of this distance being more than 31000 km is 0.6915 . Find the mean and standard deviation of the distribution.
\item The random variable $X$ has the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$, where $3 \sigma = 4 \mu$ and $\mu \neq 0$. Find $\mathrm { P } ( X < 3 \mu )$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2018 Q4 [8]}}