| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | March |
| Marks | 11 |
| 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: finding μ from a percentile (using inverse normal), calculating probabilities with binomial-normal combination, and working with a linear relationship μ = 3σ. All parts use routine methods with no novel insight required, making it slightly easier than average for S1. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = -1.645\); \(-1.645 = \dfrac{0.9 - m}{0.35}\); \(m = 1.48\) | B1, M1, A1 (3) | \(\pm 1.64\) to \(1.65\) seen; standardising with a \(z\)-value, accept \((0.35)^2\); correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Prob} = (0.9332)^4 = 0.758\) | M1, M1, A1, M1, A1 (5) | Standardising, no sq, FT their \(m\), no cc; correct area i.e. F, accept correct to 2sf; power of 4 from attempt at \(P(z)\); correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(t > 0.6\mu) = P\!\left(z > \dfrac{0.6\mu - \mu}{\mu/3}\right) = P(z > -1.2) = 0.885\) | M1, M1, A1 (3) | Standardising attempt with 1 or 2 variables; eliminating \(\mu\) or \(\sigma\); correct final answer |
## Question 7:
**(i)**
$z = -1.645$; $-1.645 = \dfrac{0.9 - m}{0.35}$; $m = 1.48$ | B1, M1, A1 (3) | $\pm 1.64$ to $1.65$ seen; standardising with a $z$-value, accept $(0.35)^2$; correct answer
**(ii)**
$P(<2) = P\!\left(z < \dfrac{2 - 1.476}{0.35}\right) = P(z < 1.50) = 0.933$
$\text{Prob} = (0.9332)^4 = 0.758$ | M1, M1, A1, M1, A1 (5) | Standardising, no sq, FT their $m$, no cc; correct area i.e. F, accept correct to 2sf; power of 4 from attempt at $P(z)$; correct answer
**(iii)**
$P(t > 0.6\mu) = P\!\left(z > \dfrac{0.6\mu - \mu}{\mu/3}\right) = P(z > -1.2) = 0.885$ | M1, M1, A1 (3) | Standardising attempt with 1 or 2 variables; eliminating $\mu$ or $\sigma$; correct final answer7 The times taken by a garage to fit a tow bar onto a car have a normal distribution with mean $m$ hours and standard deviation 0.35 hours. It is found that $95 \%$ of times taken are longer than 0.9 hours.\\
(i) Find the value of $m$.\\
(ii) On one day 4 cars have a tow bar fitted. Find the probability that none of them takes more than 2 hours to fit.
The times in hours taken by another garage to fit a tow bar onto a car have the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$ where $\mu = 3 \sigma$.\\
(iii) Find the probability that it takes more than $0.6 \mu$ hours to fit a tow bar onto a randomly chosen car at this garage.
\hfill \mbox{\textit{CAIE S1 2016 Q7 [11]}}