| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Sketch or interpret PDF graph |
| Difficulty | Moderate -0.3 This is a straightforward S2 question requiring standard techniques: sketching a simple quadratic PDF, calculating mean and variance using given formulas (with 'show that' guidance), and evaluating a definite integral. All steps are routine applications of A-level statistics methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| curve through \((1,0)\) and \((2, 2.25)\) | B1, B1 | 1 and 2 seen, no f vals needed, correct shape. |
| (ii) \(\mu = \int_1^2 [3x^3/4 - 3x/4] dx = [3x^4/16 - 3x^2/8]_1^2 = [3-12/8] - [3/16 - 3/8] = 27/16\) (1.6875) | M1, M1, A1 | attempt to integrate xf(x), any limits. correct limits on an integration. Correct answer legit obtained. |
| Answer | Marks | Guidance |
|---|---|---|
| sd = 0.2288 to 4 dp AG | M1, A1, A1 | attempt to integrate \(x^2f(x)\), with \(- \text{mean}^2\) seen. Correct integral. Correct answer legit obtained. |
| (iv) \(\int_1^{1.9163} [3x^2/4 - 3/4] dx = [x^3/4 - 3x/4]_1^{1.9163} = 0.822\) or \(0.821\) | M1, A1, A1 | attempt to integrate f(x), with limits attempted. Correct limits (4 s f). Correct answer. |
**(i)** no values outside $1 \leq x \leq 2$
curve through $(1,0)$ and $(2, 2.25)$ | B1, B1 | 1 and 2 seen, no f vals needed, correct shape.
**(ii)** $\mu = \int_1^2 [3x^3/4 - 3x/4] dx = [3x^4/16 - 3x^2/8]_1^2 = [3-12/8] - [3/16 - 3/8] = 27/16$ (1.6875) | M1, M1, A1 | attempt to integrate xf(x), any limits. correct limits on an integration. Correct answer legit obtained.
**(iii)** $\sigma^2 = \int_1^2 [3x^4/4 - 3x^2/4] dx - \text{mean}^2 = [3x^5/20 - 3x^3/12]_1^2 - 1.6875^2 = [96/20 - 2] - [3/20 - 3/12] - 1.6875^2 = 0.052343$
sd = 0.2288 to 4 dp AG | M1, A1, A1 | attempt to integrate $x^2f(x)$, with $- \text{mean}^2$ seen. Correct integral. Correct answer legit obtained.
**(iv)** $\int_1^{1.9163} [3x^2/4 - 3/4] dx = [x^3/4 - 3x/4]_1^{1.9163} = 0.822$ or $0.821$ | M1, A1, A1 | attempt to integrate f(x), with limits attempted. Correct limits (4 s f). Correct answer.
---7 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 3 } { 4 } \left( x ^ { 2 } - 1 \right) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
(i) Sketch the probability density function of $X$.\\
(ii) Show that the mean, $\mu$, of $X$ is 1.6875 .\\
(iii) Show that the standard deviation, $\sigma$, of $X$ is 0.2288 , correct to 4 decimal places.\\
(iv) Find $\mathrm { P } ( 1 \leqslant X \leqslant \mu + \sigma )$.
\hfill \mbox{\textit{CAIE S2 2007 Q7 [11]}}