| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.3 This question requires standardizing normal distributions and using z-tables, with the added step of solving simultaneous equations involving μ and σ. Part (a) is slightly above routine as it requires algebraic manipulation of the relationship σ = 0.25μ, but the techniques are standard. Parts (b) is a straightforward application of normal probability to find expected values. Overall, this is a typical S1 question with one moderately challenging element. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 0.674\) | B1 | Rounding to \(\pm 0.674\) or \(0.675\) |
| \(0.674 = \frac{6.8 - \mu}{0.25\mu}\) | M1 | Standardising, no cc, no sq rt, no sq, \(\sigma\) may still be present on RHS |
| M1 | Subst and sensible solving for \(\mu\); must collect terms, z-value needed can be 0.75 or 0.7734; need a value for \(\mu\) | |
| \(\mu = 5.82\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 4.7) = P\!\left(z < \frac{4.7 - 5.819}{1.4548}\right)\) | M1 | \(\pm\) standardising, no cc, no sq rt, no sq unless penalised in (a)(i) |
| \(= \phi(-0.769) = 1 - 0.7791\) | M1 | Correct side for their mean i.e. \(1 - \phi\) (final solution) |
| \(= 0.221\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(<15.75) = P\!\left(z < \frac{15.75 - 16}{0.2}\right) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056\) and \(P(>16.25) = 0.1056\) by symmetry | *M1 | Standardising for 15.75 or 16.25, no cc, no sq, no sq rt unless penalised in (a)(i) or (a)(ii) |
| \(P(\text{usable}) = 1 - 0.2112 = 0.7888\) | B1 | \(2\phi - 1\) OE for required probability (final solution) |
| Usable rods \(= 1000 \times 0.7888\) | DM1 | Multiply their probability by 1000, dependent on recognisable attempt to standardise |
| \(788\) or \(789\) | A1 |
## Question 6(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 0.674$ | B1 | Rounding to $\pm 0.674$ or $0.675$ |
| $0.674 = \frac{6.8 - \mu}{0.25\mu}$ | M1 | Standardising, no cc, no sq rt, no sq, $\sigma$ may still be present on RHS |
| | M1 | Subst and sensible solving for $\mu$; must collect terms, z-value needed can be 0.75 or 0.7734; need a value for $\mu$ |
| $\mu = 5.82$ | A1 | |
---
## Question 6(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 4.7) = P\!\left(z < \frac{4.7 - 5.819}{1.4548}\right)$ | M1 | $\pm$ standardising, no cc, no sq rt, no sq unless penalised in (a)(i) |
| $= \phi(-0.769) = 1 - 0.7791$ | M1 | Correct side for their mean i.e. $1 - \phi$ (final solution) |
| $= 0.221$ | A1 | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(<15.75) = P\!\left(z < \frac{15.75 - 16}{0.2}\right) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056$ and $P(>16.25) = 0.1056$ by symmetry | *M1 | Standardising for 15.75 or 16.25, no cc, no sq, no sq rt unless penalised in (a)(i) or (a)(ii) |
| $P(\text{usable}) = 1 - 0.2112 = 0.7888$ | B1 | $2\phi - 1$ OE for required probability (final solution) |
| Usable rods $= 1000 \times 0.7888$ | DM1 | Multiply their probability by 1000, dependent on recognisable attempt to standardise |
| $788$ or $789$ | A1 | |
---6
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$. You are given that $\sigma = 0.25 \mu$ and $\mathrm { P } ( X < 6.8 ) = 0.75$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $\mu$.
\item Find $\mathrm { P } ( X < 4.7 )$.
\end{enumerate}\item The lengths of metal rods have a normal distribution with mean 16 cm and standard deviation 0.2 cm . Rods which are shorter than 15.75 cm or longer than 16.25 cm are not usable. Find the expected number of usable rods in a batch of 1000 rods.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2017 Q6 [11]}}