| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate probability P(X in interval) |
| Difficulty | Moderate -0.3 This is a straightforward continuous probability distribution question requiring standard integration techniques. Part (i) involves a simple definite integral of a polynomial, while part (ii) requires computing E(X) and E(X²) using standard formulas—all routine S2 content with no conceptual challenges beyond applying learned procedures. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_{0.5}^{1} 3(1-x)^2\, dx = \left[\frac{3(1-x)^3}{-3}\right]_{0.5}^{1}\) | M1 | For attempt at integrating and using limits |
| A1 | Or equivalent correct integration (missing factors of 3 can still gain A1) | |
| \(= [0] - [-1](0.5)^3 = 0.125\) | A1 3 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \int_0^1 3x(1-x)^2\, dx = \int_0^1 3x - 6x^2 + 3x^3\, dx\) | M1 | For attempt at \(\int xf(x)\, dx\) with or without limits |
| \(= \left[\frac{3x^2}{2} - \frac{6x^3}{3} + \frac{3x^4}{4}\right]_0^1\) | A1 | For 2 or 3 correct parts of integral (missing factors of 3 can still gain A1) |
| \(= \frac{3}{2} - 2 + \frac{3}{4} = 0.25\) | A1 | For correct answer |
| \(\text{Var}(X) = \int_0^1 3x^2(1-x)^2\, dx - [E(X)]^2\) | M1 | For attempt at \(\int x^2 f(x)\, dx - [E(X)]^2\); i.e. \(-[E(X)]^2\) must be seen even if ignored in next line |
| \(= \int_0^1 3x^2 - 6x^3 + 3x^4\, dx - (0.25)^2\) | B1 | For 2 or 3 correct parts of integral (missing factors of 3 can still gain A1) |
| \(= \left[\frac{3x^3}{3} - \frac{6x^4}{4} + \frac{3x^5}{5}\right]_0^1 - (0.25)^2\) | ||
| \(= 0.0375\) | A1 6 | For correct answer |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{0.5}^{1} 3(1-x)^2\, dx = \left[\frac{3(1-x)^3}{-3}\right]_{0.5}^{1}$ | M1 | For attempt at integrating and using limits |
| | A1 | Or equivalent correct integration (missing factors of 3 can still gain A1) |
| $= [0] - [-1](0.5)^3 = 0.125$ | A1 **3** | For correct answer |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \int_0^1 3x(1-x)^2\, dx = \int_0^1 3x - 6x^2 + 3x^3\, dx$ | M1 | For attempt at $\int xf(x)\, dx$ with or without limits |
| $= \left[\frac{3x^2}{2} - \frac{6x^3}{3} + \frac{3x^4}{4}\right]_0^1$ | A1 | For 2 or 3 correct parts of integral (missing factors of 3 can still gain A1) |
| $= \frac{3}{2} - 2 + \frac{3}{4} = 0.25$ | A1 | For correct answer |
| $\text{Var}(X) = \int_0^1 3x^2(1-x)^2\, dx - [E(X)]^2$ | M1 | For attempt at $\int x^2 f(x)\, dx - [E(X)]^2$; i.e. $-[E(X)]^2$ must be seen even if ignored in next line |
| $= \int_0^1 3x^2 - 6x^3 + 3x^4\, dx - (0.25)^2$ | B1 | For 2 or 3 correct parts of integral (missing factors of 3 can still gain A1) |
| $= \left[\frac{3x^3}{3} - \frac{6x^4}{4} + \frac{3x^5}{5}\right]_0^1 - (0.25)^2$ | | |
| $= 0.0375$ | A1 **6** | For correct answer |
---6 A continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} 3 ( 1 - x ) ^ { 2 } & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
Find\\
(i) $\mathrm { P } ( X > 0.5 )$,\\
(ii) the mean and variance of $X$.
\hfill \mbox{\textit{CAIE S2 2004 Q6 [9]}}