| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| 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 on continuous probability distributions requiring routine techniques: sketching a piecewise pdf, using the integral property to find k (solving a quadratic), defining the CDF by integration, calculating a probability, and identifying median/mode from the graph. All parts follow 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 pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape (straight lines, all above \(x\)-axis) | B1 | |
| Correct labels: \(1\), \(k\), \(0.5\), \(k-0.5\) in correct places | B1 | Allow use of \(\frac{1}{2}(1+\sqrt{5})\)/awrt 1.62 instead of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^k \left(x - \frac{1}{2}\right)dx = \frac{1}{2}\) | M1 | |
| \(\left[\frac{1}{2}x^2 - \frac{1}{2}x\right]_1^k = \frac{1}{2}\) | ||
| \(k^2 - k - 1 = 0\) | A1 | |
| \(k = \frac{1}{2}(1 + \sqrt{5})\) | M1A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(x) = 0,\quad x < 0\) | B1 | 1st and last |
| \(F(x) = \frac{1}{2}x,\quad 0 \leq x < 1\) | B1 | |
| \(F(x) = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2},\quad 1 \leq x \leq k\) | M1A1A1B1 | Working: \(\int_1^x \left(t - \frac{1}{2}\right)dt + C = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2}\) gives (M1A1;A1) |
| \(F(x) = 1,\quad x > k\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(0.5 < X < 1.5) = F(1.5) - F(0.5)\) | M1 | |
| \(= 0.875 - 0.25 = 0.625\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Median is \(x = 1\) | B1 | |
| Mode is \(x = k\) or \(\frac{1}{2}(1+\sqrt{5})\) or awrt \(1.62\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Negative skew | B1 | |
| Median \(<\) mode or from graph more values are to the right | B1d | Dependent B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^k x - \frac{1}{2}\,dx = 0.5\) | M1 | Or \(\int_1^k x - \frac{1}{2}\,dx + 0.5 = 1\) ignore limits; or \(\frac{1}{2}(k-0.5+0.5)(k-1) = 0.5\) |
| Quadratic in form \(a(k^2 - k - 1) = 0\) or \(ak^2 - ak = a\) | A1 | Where \(a\) is a constant |
| Solving quadratic of form \(ak^2 - bk + c = 0\) | M1 | Where \(a,b,c \neq 0\); must be at least one correct step |
| \(k = \frac{1}{2}(1+\sqrt{5})\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Second line condition | B1 | Do not penalise use of \(<\) instead of \(\leq\) |
| \(\int_1^k x - \frac{1}{2}\,dx + C\), integration giving \(x \rightarrow x^2\) | M1 | Ignore limits |
| \(\frac{1}{2}x^2 - \frac{1}{2}x\) | A1 | Correct integration |
| \(C = \frac{1}{2}\) | A1 | |
| Third line: \(\frac{1}{2}x^2 - \frac{1}{2}x \pm C\) | B1 | Do not penalise \(<\) instead of \(\leq\); \(C\) may equal 0 |
| First and last line | B1 | Do not penalise \(\leq\) instead of \(<\); allow \(k\) or value of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(F(1.5) - F(0.5)\); 1.5 in third line of c.d.f., 0.5 in second line | M1 | |
| \(\int_{0.5}^1 \frac{1}{2}x\,dx + \int_1^{1.5} x - \frac{1}{2}\,dx\) | Need to attempt integration, at least one \(x^n \rightarrow x^{n+1}\) | |
| \(0.25 + 0.375\) or correct method | If \(C=0\) used, get 0.125 → M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Median then mode (if unclear which is which, assume median is first) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Negative/negative skew(ness) | B1 | Do not allow negative correlation |
| Reason following from values or diagram | B1 | Dependent on previous B mark |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape (straight lines, all above $x$-axis) | B1 | |
| Correct labels: $1$, $k$, $0.5$, $k-0.5$ in correct places | B1 | Allow use of $\frac{1}{2}(1+\sqrt{5})$/awrt 1.62 instead of $k$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^k \left(x - \frac{1}{2}\right)dx = \frac{1}{2}$ | M1 | |
| $\left[\frac{1}{2}x^2 - \frac{1}{2}x\right]_1^k = \frac{1}{2}$ | | |
| $k^2 - k - 1 = 0$ | A1 | |
| $k = \frac{1}{2}(1 + \sqrt{5})$ | M1A1 cso | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(x) = 0,\quad x < 0$ | B1 | 1st and last |
| $F(x) = \frac{1}{2}x,\quad 0 \leq x < 1$ | B1 | |
| $F(x) = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2},\quad 1 \leq x \leq k$ | M1A1A1B1 | Working: $\int_1^x \left(t - \frac{1}{2}\right)dt + C = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2}$ gives (M1A1;A1) |
| $F(x) = 1,\quad x > k$ | B1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(0.5 < X < 1.5) = F(1.5) - F(0.5)$ | M1 | |
| $= 0.875 - 0.25 = 0.625$ | A1 | |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Median is $x = 1$ | B1 | |
| Mode is $x = k$ or $\frac{1}{2}(1+\sqrt{5})$ or awrt $1.62$ | B1 | |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Negative skew | B1 | |
| Median $<$ mode or from graph more values are to the right | B1d | Dependent B1 |
# Question (b) [Probability Distribution]:
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^k x - \frac{1}{2}\,dx = 0.5$ | M1 | Or $\int_1^k x - \frac{1}{2}\,dx + 0.5 = 1$ ignore limits; or $\frac{1}{2}(k-0.5+0.5)(k-1) = 0.5$ |
| Quadratic in form $a(k^2 - k - 1) = 0$ or $ak^2 - ak = a$ | A1 | Where $a$ is a constant |
| Solving quadratic of form $ak^2 - bk + c = 0$ | M1 | Where $a,b,c \neq 0$; must be at least one correct step |
| $k = \frac{1}{2}(1+\sqrt{5})$ | A1 | cso |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Second line condition | B1 | Do not penalise use of $<$ instead of $\leq$ |
| $\int_1^k x - \frac{1}{2}\,dx + C$, integration giving $x \rightarrow x^2$ | M1 | Ignore limits |
| $\frac{1}{2}x^2 - \frac{1}{2}x$ | A1 | Correct integration |
| $C = \frac{1}{2}$ | A1 | |
| Third line: $\frac{1}{2}x^2 - \frac{1}{2}x \pm C$ | B1 | Do not penalise $<$ instead of $\leq$; $C$ may equal 0 |
| First and last line | B1 | Do not penalise $\leq$ instead of $<$; allow $k$ or value of $k$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $F(1.5) - F(0.5)$; 1.5 in third line of c.d.f., 0.5 in second line | M1 | |
| $\int_{0.5}^1 \frac{1}{2}x\,dx + \int_1^{1.5} x - \frac{1}{2}\,dx$ | | Need to attempt integration, at least one $x^n \rightarrow x^{n+1}$ |
| $0.25 + 0.375$ or correct method | | If $C=0$ used, get 0.125 → M1A0 |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Median then mode (if unclear which is which, assume median is first) | | |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Negative/negative skew(ness) | B1 | Do not allow negative correlation |
| Reason following from values or diagram | B1 | Dependent on previous B mark |
---6. A random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 0 \leqslant x < 1 \\ x - \frac { 1 } { 2 } & 1 \leqslant x \leqslant k \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $\mathrm { f } ( x )$.
\item Show that $k = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )$.
\item Define fully the cumulative distribution function $\mathrm { F } ( x )$.
\item Find $\mathrm { P } ( 0.5 < X < 1.5 )$.
\item Write down the median of $X$ and the mode of $X$.
\item Describe the skewness of the distribution of $X$. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2012 Q6 [18]}}