| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| 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 continuous probability distribution question requiring routine techniques: sketching a piecewise pdf, identifying the mode by inspection, using the normalization condition to find k, and basic probability calculations. The skewness comparison (mean vs mode) and quartile identification require understanding but minimal computation. While multi-part, each step follows textbook 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 integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal line \(y=3k\) with \(3k\) marked on \(y\)-axis | B1 | 1st B1 for horizontal line \(y=3k\) and \(3k\) marked on \(y\)-axis |
| Correct curve shape for \(1 < x < 4\), meeting \(x\)-axis at \((4,0)\), not extending below \(x\)-axis. Must be a curve | B1 | 2nd B1 for correct shape |
| \(x=1\) marked and graphs meeting at point \((1, 3k)\) | B1 (3) | 3rd B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mode \(= 2\) | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(<\) mode, so negative skew | B1, dB1 (2) | 1st B1 for suitable reason matching their mode (mode must be a number, must use mean). 2nd dB1 not ft, dependent on 1st B1. Correct answer from correct value of mode |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3k \times 1 + \int_1^4 (4kx - kx^2)\,dx = 1\) | M1, B1 | 1st M1 for attempting sum of both areas \(=1\), ignore limits. B1 for \(3k\) seen added to integral |
| \(3k + \left[2kx^2 - \frac{kx^3}{3}\right]_1^4 \{=1\}\) | M1 | 2nd M1 for some correct integration, at least one \(kx^n \to kx^{n+1}\) |
| \(3k + \left(32k - \frac{64k}{3}\right) - \left(2k - \frac{k}{3}\right) = 1\) | M1d | 3rd M1d dependent on 1st M1. For use of correct limits |
| \(12k = 1\) so \(k = \frac{1}{12}\) | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Lower Quartile \(= 1\) | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(1 < X < 2) = P(2 < X < 3)\) by symmetry | M1 | M1 for identifying the symmetry, may be implied by \(P(1{<}x{<}2) = \frac{11}{36}\) found by any method, or writing a correct equation (ft their \(k\)) |
| So \(P(X>3) = 1 - 3k - \frac{22}{36} = \frac{5}{36}\) | A1 (2) | A1 for \(\frac{5}{36}\) or exact equivalent |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal line $y=3k$ with $3k$ marked on $y$-axis | B1 | 1st B1 for horizontal line $y=3k$ and $3k$ marked on $y$-axis |
| Correct curve shape for $1 < x < 4$, meeting $x$-axis at $(4,0)$, not extending below $x$-axis. Must be a curve | B1 | 2nd B1 for correct shape |
| $x=1$ marked and graphs meeting at point $(1, 3k)$ | B1 (3) | 3rd B1 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mode $= 2$ | B1 (1) | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $<$ mode, so negative skew | B1, dB1 (2) | 1st B1 for suitable reason matching their mode (mode must be a number, must use mean). 2nd dB1 not ft, dependent on 1st B1. Correct answer from correct value of mode |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3k \times 1 + \int_1^4 (4kx - kx^2)\,dx = 1$ | M1, B1 | 1st M1 for attempting sum of both areas $=1$, ignore limits. B1 for $3k$ seen added to integral |
| $3k + \left[2kx^2 - \frac{kx^3}{3}\right]_1^4 \{=1\}$ | M1 | 2nd M1 for some correct integration, at least one $kx^n \to kx^{n+1}$ |
| $3k + \left(32k - \frac{64k}{3}\right) - \left(2k - \frac{k}{3}\right) = 1$ | M1d | 3rd M1d dependent on 1st M1. For use of correct limits |
| $12k = 1$ so $k = \frac{1}{12}$ | A1 (5) | |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Lower Quartile $= 1$ | B1 (1) | |
## Part (f)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(1 < X < 2) = P(2 < X < 3)$ by symmetry | M1 | M1 for identifying the symmetry, may be implied by $P(1{<}x{<}2) = \frac{11}{36}$ found by any method, or writing a correct equation (ft their $k$) |
| So $P(X>3) = 1 - 3k - \frac{22}{36} = \frac{5}{36}$ | A1 (2) | A1 for $\frac{5}{36}$ or exact equivalent |
---4. The random variable $X$ has probability density function $\mathrm { f } ( x )$ given by
$$\mathrm { f } ( x ) = \left\{ \begin{array} { c c }
3 k & 0 \leqslant x < 1 \\
k x ( 4 - x ) & 1 \leqslant x \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch f (x).
\item Write down the mode of $X$.
Given that $\mathrm { E } ( X ) = \frac { 29 } { 16 }$
\item describe, giving a reason, the skewness of the distribution.
\item Use integration to find the value of $k$.
\item Write down the lower quartile of $X$.
Given also that $\mathrm { P } ( 2 < X < 3 ) = \frac { 11 } { 36 }$
\item find the exact value of $\mathrm { P } ( X > 3 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q4 [14]}}