| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Compare mean and median using probability |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques and knowledge of pdf properties. Part (a) uses the integral-equals-1 property, parts (b-d) involve straightforward integration for E(X) and CDF, and part (e) is simple interpretation. All steps are textbook exercises 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 integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | \(\int (ax^2 + bx) \, dx = [1]\) | M1 |
| \(\left[\frac{ax^3}{3} + \frac{bx^2}{2}\right] = [1]\) | A1 | 1st A1 fully correct integration. Ignore limits and accept any letters. |
| \(\left(\frac{a(7^3)}{3} + \frac{b(7^2)}{2}\right) - \left(\frac{a}{3} + \frac{b}{2}\right) = 1\) | dM1 | 2nd dM1 dep on 1st M1. Subst in correct limits and equate to 1 |
| \(114a + 24b = 1\) \(*\) | A1 cso | 3rd A1 correct solution only (answer given) |
| Part (b) | \(114(\frac{1}{30}) + 24b = 1 \Rightarrow b = -\frac{1}{90}\) | B1 |
| \([E(X)] = \int ({\frac{1}{30}x^3 + b'x^2}) \, dx\) | M1 | 1st M1 attempting to integrate xf(x), (at least one term \(x^n \to x^{n+1}\)). Ignore limits. |
| \(\left[\frac{x^4}{360} - \frac{x^3}{270}\right] = \left(\frac{7^4}{360} - \frac{7^3}{270}\right) - \left(\frac{1}{360} - \frac{1}{270}\right) = 5.4\) | A1ft, A1oe | 1st A1ft fully correct integration ft their value of \(b\). (Allow if given in terms of \(a\) and \(b\)); 2nd A1 5.4 oe |
| Part (c) | \(\int_{1}^{\dot{t}} (\frac{1}{30}t^2 - \frac{1}{90}t) \, dt = [\frac{1}{270} - \frac{1}{180}]\) or \(\int (\frac{1}{30}x^2 - \frac{1}{90}x) \, dx = \frac{x^3}{270} - \frac{x^2}{180} + c\) with F(1) = 0 or F(7) = 1 | M1 |
| \(F(x) = \begin{cases} 0 & x < 1 \\ \frac{x^3}{270} - \frac{x^2}{180} + \frac{1}{540} & 1 \leq x \leq 7 \\ 1 & x > 7 \end{cases}\) | A1 B1 | A1 \(\frac{x^3}{270} - \frac{x^2}{180} + \frac{1}{540}\) with correct limits (allow \(< \) or \(\leq\)); B1 for top and bottom lines with correct limits |
| Part (d) | \(P(X > 5.4') = 1 - F(5.4') = 1 - 0.42305\ldots = 0.5769\ldots\) | M1 A1 |
| Part (e) | Since (d) > 0.5, [the mean is less than the median] therefore negative (skew). | M1, A1 |
| Total (15) |
| **Part (a)** | $\int (ax^2 + bx) \, dx = [1]$ | M1 | 1st M1 attempting to integrate f(x), (at least one term $x^n \to x^{n+1}$). Ignore limits. |
| | $\left[\frac{ax^3}{3} + \frac{bx^2}{2}\right] = [1]$ | A1 | 1st A1 fully correct integration. Ignore limits and accept any letters. |
| | $\left(\frac{a(7^3)}{3} + \frac{b(7^2)}{2}\right) - \left(\frac{a}{3} + \frac{b}{2}\right) = 1$ | dM1 | 2nd dM1 dep on 1st M1. Subst in correct limits and equate to 1 |
| | $114a + 24b = 1$ $*$ | A1 cso | 3rd A1 correct solution only (answer given) |
| **Part (b)** | $114(\frac{1}{30}) + 24b = 1 \Rightarrow b = -\frac{1}{90}$ | B1 | |
| | $[E(X)] = \int ({\frac{1}{30}x^3 + b'x^2}) \, dx$ | M1 | 1st M1 attempting to integrate xf(x), (at least one term $x^n \to x^{n+1}$). Ignore limits. |
| | $\left[\frac{x^4}{360} - \frac{x^3}{270}\right] = \left(\frac{7^4}{360} - \frac{7^3}{270}\right) - \left(\frac{1}{360} - \frac{1}{270}\right) = 5.4$ | A1ft, A1oe | 1st A1ft fully correct integration ft their value of $b$. (Allow if given in terms of $a$ and $b$); 2nd A1 5.4 oe |
| **Part (c)** | $\int_{1}^{\dot{t}} (\frac{1}{30}t^2 - \frac{1}{90}t) \, dt = [\frac{1}{270} - \frac{1}{180}]$ or $\int (\frac{1}{30}x^2 - \frac{1}{90}x) \, dx = \frac{x^3}{270} - \frac{x^2}{180} + c$ with F(1) = 0 or F(7) = 1 | M1 | M1 attempt to integrate f(x) with correct limits or with $+ C$ and attempt at F(1) = 0 or F(7) = 1 |
| | $F(x) = \begin{cases} 0 & x < 1 \\ \frac{x^3}{270} - \frac{x^2}{180} + \frac{1}{540} & 1 \leq x \leq 7 \\ 1 & x > 7 \end{cases}$ | A1 B1 | A1 $\frac{x^3}{270} - \frac{x^2}{180} + \frac{1}{540}$ with correct limits (allow $< $ or $\leq$); B1 for top and bottom lines with correct limits |
| **Part (d)** | $P(X > 5.4') = 1 - F(5.4') = 1 - 0.42305\ldots = 0.5769\ldots$ | M1 A1 | M1 $1 - F(5.4')$ or $F(7) - F(5.4')$ or $\int_{5.4'}^{7} (\frac{1}{30}x^2 + (-\frac{1}{90}'x)) \, dx$; awrt **0.577** |
| **Part (e)** | Since (d) > 0.5, [the mean is less than the median] therefore negative (skew). | M1, A1 | M1 for correctly comparing 'their (d)' with 0.5 (may be implied by a correct comparison of mean and median ft their (d)). If no answer given in (d), then M0. A1 for negative skew which must follow from 'their (d)' $>$ 0.5 |
| | | | Total (15) |
---A continuous random variable $X$ has probability density function
$$\mathrm{f}(x) = \begin{cases}
ax^2 + bx & 1 \leq x \leq 7 \\
0 & \text{otherwise}
\end{cases}$$
where $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Show that $114a + 24b = 1$ [4]
\end{enumerate}
Given that $a = \frac{1}{90}$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item use algebraic integration to find $\mathrm{E}(X)$ [4]
\item find the cumulative distribution function of $X$, specifying it for all values of $x$ [3]
\item find $\mathrm{P}(X > \mathrm{E}(X))$ [2]
\item use your answer to part (d) to describe the skewness of the distribution. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q6 [15]}}