| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sample statistics from uniform |
| Difficulty | Moderate -0.5 This is a straightforward application of uniform distribution properties with standard formulas for mean (b/2) and variance (b²/12), followed by routine calculations using the sample mean to estimate b, and applying normal approximation to the sampling distribution. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(E(X) = \int_0^b x \cdot f(x) \, dx = \left[\frac{x^2}{2b}\right]_0^b = \frac{b}{2}\) | B1 | Correct answer (accept unsimplified) |
| \(\text{Var}(X) = \int_0^b \frac{x^2}{b} \, dx - \frac{b^2}{4} = \frac{b^2}{12}\) | M1 | For (substituted) attempt at \(\int x^2 f(x) \, dx - [E(X)^2]\) ie \(-[E(X^2)]\) must be seen even if ignored in next line |
| A1 | 3 | Correct answer. Accept unsimplified – but must be a single fraction. |
| (ii) \(9.5 = b/2\) | M1 | Equating their mean to their 9.5 |
| \(b = 19\) AG | A1 | 2 |
| (iii) \(8/19\) or \(0.421\) | B1 | 1 |
| (iv) \(\bar{X} \sim N(9.5, 30.08/336)\) or using totals \(N(3192, 10106.88)\) | M1 | Dividing their \(b^2/12\) by 336 |
| A1 ft | Correct mean and variance | |
| \(P(\bar{X} < 9) = P\left(z < \frac{9 - 9.5}{\sqrt{30.08/336}}\right)\) or equiv | M1 | Standardising (must involve 336) and area \(< 0.5\) or consistent with their figures |
| \(= P(z < -1.671) = 1 - 0.9526 = 0.0474\) | A1 | 4 |
**(i)** $E(X) = \int_0^b x \cdot f(x) \, dx = \left[\frac{x^2}{2b}\right]_0^b = \frac{b}{2}$ | B1 | Correct answer (accept unsimplified)
$\text{Var}(X) = \int_0^b \frac{x^2}{b} \, dx - \frac{b^2}{4} = \frac{b^2}{12}$ | M1 | For (substituted) attempt at $\int x^2 f(x) \, dx - [E(X)^2]$ ie $-[E(X^2)]$ must be seen even if ignored in next line
| A1 | 3 | Correct answer. Accept unsimplified – but must be a **single fraction**.
**(ii)** $9.5 = b/2$ | M1 | Equating their mean to their 9.5
$b = 19$ AG | A1 | 2 | Correct answer
**(iii)** $8/19$ or $0.421$ | B1 | 1 | Correct answer
**(iv)** $\bar{X} \sim N(9.5, 30.08/336)$ or using totals $N(3192, 10106.88)$ | M1 | Dividing their $b^2/12$ by 336
| A1 ft | Correct mean and variance
$P(\bar{X} < 9) = P\left(z < \frac{9 - 9.5}{\sqrt{30.08/336}}\right)$ or equiv | M1 | Standardising (must involve 336) and area $< 0.5$ or consistent with their figures
$= P(z < -1.671) = 1 - 0.9526 = 0.0474$ | A1 | 4 | Correct answer
---5 The length, $X \mathrm {~cm}$, of a piece of wooden planking is a random variable with probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { b } & 0 \leqslant x \leqslant b \\ 0 & \text { otherwise } \end{cases}$$
where $b$ is a positive constant.\\
(i) Find the mean and variance of $X$ in terms of $b$.
The lengths of a random sample of 100 pieces were measured and it was found that $\Sigma x = 950$.\\
(ii) Show that the value of $b$ estimated from this information is 19 .
Using this value of $b$,\\
(iii) find the probability that the length of a randomly chosen piece is greater than 11 cm ,\\
(iv) find the probability that the mean length of a random sample of 336 pieces is less than 9 cm .
\hfill \mbox{\textit{CAIE S2 2007 Q5 [10]}}