| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Moderate -0.3 This is a standard S2 question testing routine CDF/PDF manipulation. Part (a) requires direct CDF evaluation, (b) is straightforward differentiation (marked as 'show that'), (c) involves standard expectation/variance integration formulas, and (d) applies linear transformation rules. All techniques are textbook exercises with no problem-solving insight 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 integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(0.4 \leq X \leq 0.8) = F(0.8) - F(0.4)\) | M1 | Correct use of CDF |
| \(F(0.8) = \frac{1}{2}(0.8) - \frac{1}{16}(0.8)^2 = 0.4 - 0.04 = 0.36\) | ||
| \(F(0.4) = \frac{1}{2}(0.4) - \frac{1}{16}(0.4)^2 = 0.2 - 0.01 = 0.19\) | ||
| \(P = 0.36 - 0.19 = 0.17\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(F(x)\): \(f(x) = \frac{d}{dx}\left(\frac{1}{2}x - \frac{1}{16}x^2\right) = \frac{1}{2} - \frac{1}{8}x\) for \(0 \leq x \leq 4\) | B1 | Correct differentiation shown, with correct limits and \(f(x)=0\) otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \int_0^4 x\left(\frac{1}{2} - \frac{1}{8}x\right)dx\) | M1 | Correct integral set up |
| \(= \int_0^4 \left(\frac{x}{2} - \frac{x^2}{8}\right)dx = \left[\frac{x^2}{4} - \frac{x^3}{24}\right]_0^4\) | A1 | Correct integration |
| \(= \frac{16}{4} - \frac{64}{24} = 4 - \frac{8}{3} = \frac{4}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = \int_0^4 x^2\left(\frac{1}{2} - \frac{1}{8}x\right)dx\) | M1 | |
| \(= \left[\frac{x^3}{6} - \frac{x^4}{32}\right]_0^4 = \frac{64}{6} - \frac{256}{32} = \frac{32}{3} - 8 = \frac{8}{3}\) | A1 | |
| \(\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{8}{3} - \frac{16}{9} = \frac{24-16}{9} = \frac{8}{9}\) | M1 A1 | Correct use of \(\text{Var}(X)=E(X^2)-\mu^2\); correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = 3E(X) - 2 = 3\times\frac{4}{3} - 2 = 2\) | B1 | |
| \(\text{Var}(Y) = 9\,\text{Var}(X) = 9 \times \frac{8}{9} = 8\) | B1 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(0.4 \leq X \leq 0.8) = F(0.8) - F(0.4)$ | M1 | Correct use of CDF |
| $F(0.8) = \frac{1}{2}(0.8) - \frac{1}{16}(0.8)^2 = 0.4 - 0.04 = 0.36$ | | |
| $F(0.4) = \frac{1}{2}(0.4) - \frac{1}{16}(0.4)^2 = 0.2 - 0.01 = 0.19$ | | |
| $P = 0.36 - 0.19 = 0.17$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $F(x)$: $f(x) = \frac{d}{dx}\left(\frac{1}{2}x - \frac{1}{16}x^2\right) = \frac{1}{2} - \frac{1}{8}x$ for $0 \leq x \leq 4$ | B1 | Correct differentiation shown, with correct limits and $f(x)=0$ otherwise |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \int_0^4 x\left(\frac{1}{2} - \frac{1}{8}x\right)dx$ | M1 | Correct integral set up |
| $= \int_0^4 \left(\frac{x}{2} - \frac{x^2}{8}\right)dx = \left[\frac{x^2}{4} - \frac{x^3}{24}\right]_0^4$ | A1 | Correct integration |
| $= \frac{16}{4} - \frac{64}{24} = 4 - \frac{8}{3} = \frac{4}{3}$ | A1 | |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = \int_0^4 x^2\left(\frac{1}{2} - \frac{1}{8}x\right)dx$ | M1 | |
| $= \left[\frac{x^3}{6} - \frac{x^4}{32}\right]_0^4 = \frac{64}{6} - \frac{256}{32} = \frac{32}{3} - 8 = \frac{8}{3}$ | A1 | |
| $\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{8}{3} - \frac{16}{9} = \frac{24-16}{9} = \frac{8}{9}$ | M1 A1 | Correct use of $\text{Var}(X)=E(X^2)-\mu^2$; correct conclusion |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = 3E(X) - 2 = 3\times\frac{4}{3} - 2 = 2$ | B1 | |
| $\text{Var}(Y) = 9\,\text{Var}(X) = 9 \times \frac{8}{9} = 8$ | B1 | |6 The continuous random variable $X$ has the cumulative distribution function
$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 2 } x - \frac { 1 } { 16 } x ^ { 2 } & 0 \leqslant x \leqslant 4 \\ 1 & x > 4 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability that $X$ lies between 0.4 and 0.8 .
\item Show that the probability density function, $\mathrm { f } ( x )$, of $X$ is given by
$$f ( x ) = \begin{cases} \frac { 1 } { 2 } - \frac { 1 } { 8 } x & 0 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\mathrm { E } ( X )$.
\item Show that $\operatorname { Var } ( X ) = \frac { 8 } { 9 }$.
\end{enumerate}\item The continuous random variable $Y$ is defined by
$$Y = 3 X - 2$$
Find the values of $\mathrm { E } ( Y )$ and $\operatorname { Var } ( Y )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2015 Q6 [12]}}