| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine integration to find k, construct the CDF piecewise, calculate E(X), and verify the median location. All techniques are straightforward applications of definitions with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03d E(g(X)): general expectation formula5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\int_0^3 k(x^2 - 2x + 2)dx + \int_3^4 3kdx = 1\) | M1 | \(k\left[\frac{1}{3}x^3 - x^2 + 2x\right]_0^3 + [3kx]_3^4 = 1\) or \(k\left(\frac{1}{3}x^3 - x^2 + 2x\right)\Big |
| (b) For \(0 < x \leq 3\): \(F(x) = \int_0^x \frac{1}{9}(t^2 - 2t + 2)dt = \frac{1}{9}\left(\frac{1}{3}x^3 - x^2 + 2x\right)\) | M1 A1 | For \(3 < x \leq 4\): \(F(x) = \int_3^x 3kdt + \frac{2}{3} = \frac{x}{3} - 1\) |
| (c) \(E(X) = \int_0^3 \frac{x}{9}(x^2 - 2x + 2)dt + \int_3^4 \frac{x}{3}dx = \frac{1}{9}\left[\frac{1}{4}x^4 - \frac{2}{3}x^3 + x^2\right]_0^3 + \left[\frac{1}{6}x^2\right]_3^4 = \frac{29}{12}\) or \(2.416\) or awrt \(2.42\) | M1 A1 A1 | (3 marks) |
| (d) \(F(m) = 0.5\) | M1 | \(F(2.6) = \frac{1}{27}(2.6^3 - 3 \times 2.6^2 + 6 \times 2.6) = \text{awrt } 0.48\) |
| Total [17] |
**(a)** $\int_0^3 k(x^2 - 2x + 2)dx + \int_3^4 3kdx = 1$ | M1 | $k\left[\frac{1}{3}x^3 - x^2 + 2x\right]_0^3 + [3kx]_3^4 = 1$ or $k\left(\frac{1}{3}x^3 - x^2 + 2x\right)\Big|_0^3 + 3k\Big|_0^4 = 1$ | A1 M1 dep | $9k = 1$, so $k = \frac{1}{9}$ **given** | cso A1 | (4 marks)
**(b)** For $0 < x \leq 3$: $F(x) = \int_0^x \frac{1}{9}(t^2 - 2t + 2)dt = \frac{1}{9}\left(\frac{1}{3}x^3 - x^2 + 2x\right)$ | M1 A1 | For $3 < x \leq 4$: $F(x) = \int_3^x 3kdt + \frac{2}{3} = \frac{x}{3} - 1$ | M1 A1 | $$F(x) = \begin{cases} 0 & x \leq 0 \\ \frac{1}{27}(x^3 - 3x^2 + 6x) & 0 < x \leq 3 \\ \frac{x}{3} - \frac{1}{3} & 3 < x \leq 4 \\ 1 & x > 4 \end{cases}$$ | B1 ft B1 | (6 marks)
**(c)** $E(X) = \int_0^3 \frac{x}{9}(x^2 - 2x + 2)dt + \int_3^4 \frac{x}{3}dx = \frac{1}{9}\left[\frac{1}{4}x^4 - \frac{2}{3}x^3 + x^2\right]_0^3 + \left[\frac{1}{6}x^2\right]_3^4 = \frac{29}{12}$ or $2.416$ or awrt $2.42$ | M1 A1 A1 | (3 marks)
**(d)** $F(m) = 0.5$ | M1 | $F(2.6) = \frac{1}{27}(2.6^3 - 3 \times 2.6^2 + 6 \times 2.6) = \text{awrt } 0.48$ | M1 | $F(2.7) = \frac{1}{27}(2.7^3 - 3 \times 2.7^2 + 6 \times 2.7) = \text{awrt } 0.52$ | A1 | Hence median lies between 2.6 and 2.7 | A1 dA | (4 marks)
| **Total [17]**
---\begin{enumerate}
\item The continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ given by
\end{enumerate}
$$f ( x ) = \begin{cases} k \left( x ^ { 2 } - 2 x + 2 \right) & 0 < x \leqslant 3 \\ 3 k & 3 < x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(a) Show that $k = \frac { 1 } { 9 }$\\
(b) Find the cumulative distribution function $\mathrm { F } ( x )$.\\
(c) Find the mean of $X$.\\
(d) Show that the median of $X$ lies between $x = 2.6$ and $x = 2.7$
\hfill \mbox{\textit{Edexcel S2 2010 Q4 [17]}}