| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard techniques: part (i) requires finding P(1.82 < X < 1.92) and raising to the 4th power for independence; part (ii) is a reverse normal calculation using symmetry and tables. Both are routine S1 procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For standardising one value, no cc | M1 | |
| For standardising the other value, no cc. SR ft on no sq rt | M1 | |
| For finding correct area (i.e. two \(\Phi\)s - 1) | M1 | |
| For correct answer | A1 | |
| For correct answer, ft from their (i). if \(p<1\), allow 0.0043 | A1ft [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For correct \(z\), + or - or both | B1 | |
| For seeing an equation involving + or - of their \(z\), \(b\), 0.15 (their \(z\) can only be 0.842 or 0.84 or 0.841) | M1 | |
| both limits needed, ft 1.77 to 2.03 on 0.842 only | A1ft [3] |
(i) $z_1 = \frac{0.02}{0.15} = 0.1333$
$z_2 = -\frac{0.08}{0.15} = -0.5333$
area $= \Phi(0.1333) - \Phi(-0.533) = \Phi(0.1333) - [1 - \Phi(0.5333)] = 0.5529 + 0.7029 - 1 = 0.256$
| Answer | Marks | Guidance |
|--------|-------|----------|
| For standardising one value, no cc | M1 | |
| For standardising the other value, no cc. SR ft on no sq rt | M1 | |
| For finding correct area (i.e. two $\Phi$s - 1) | M1 | |
| For correct answer | A1 | |
| For correct answer, ft from their (i). if $p<1$, allow 0.0043 | A1ft [5] | |
(ii) $z = \pm 1.282$ or 1.28 or 1.281
$\pm 1.282 = \frac{b}{0.15}$
limits between 1.71 and 2.09
| Answer | Marks | Guidance |
|--------|-------|----------|
| For correct $z$, + or - or both | B1 | |
| For seeing an equation involving + or - of their $z$, $b$, 0.15 (their $z$ can only be 0.842 or 0.84 or 0.841) | M1 | |
| both limits needed, ft 1.77 to 2.03 on 0.842 only | A1ft [3] | |
---6 Tyre pressures on a certain type of car independently follow a normal distribution with mean 1.9 bars and standard deviation 0.15 bars.\\
(i) Find the probability that all four tyres on a car of this type have pressures between 1.82 bars and 1.92 bars.\\
(ii) Safety regulations state that the pressures must be between $1.9 - b$ bars and $1.9 + b$ bars. It is known that $80 \%$ of tyres are within these safety limits. Find the safety limits.
\hfill \mbox{\textit{CAIE S1 2005 Q6 [8]}}