| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Standard +0.3 This question involves standard formula application for continuous uniform distributions (variance formula) and straightforward expectation calculations using linearity. Part (a) is algebraic manipulation of the variance formula, while part (b) requires computing E(X²), E(X), and E(Y) then combining them—all routine techniques for FS2 with no novel problem-solving required. |
| Spec | 5.02e Discrete uniform distribution5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{12}(a-5)^2 = \frac{27}{4}\) | M1 | 3.1a — translating problem into correct equation; allow \(\frac{a^3-125}{3(a-5)} - \left(\frac{a+5}{2}\right)^2 = \frac{27}{4}\) |
| \((a-5)^2 = 81\), so \(a - 5 = 9\) or \(a - 5 = -9\) | A1 | 1.1b — for \(a-5=9\) or \(a-5=-9\) or equivalent quadratic/cubic |
| Since \(a > 5\), \(a = 14\) | A1cso* | 2.2a — concluding \(a=14\) giving reason why \(-4\) is rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct method for \(\text{E}(Y)\), \(\text{E}(X)\) and \(\text{E}(X^2)\) or \(\text{E}(Y)\) and \(\text{E}(X^2+X)\) | M1 | 3.1a — complete method to solve the problem |
| \(\text{E}(Y) = \int_2^6 \frac{1}{20} y(2y-3)\, dy\) | M1 | 1.1b — attempt at \(\text{E}(Y)\) |
| \(= \frac{68}{15}\) | A1 | 1.1b — awrt 4.53 |
| \(\text{E}(X) = \frac{5+14}{2} = 9.5\) and \(\frac{27}{4} = \text{E}(X^2) - 9.5^2\), or \(\int_5^{14} \frac{x^2}{9}dx\) or \(\int_5^{14}\left(\frac{x^3}{9}+\frac{x^2}{9}\right)dx\) | M1 | 1.1b — attempt at \(\text{E}(X)\) and \(\text{E}(X^2)\) or \(\text{E}(X^2+X)\); allow \(\text{Var}(X)=\text{E}(X^2)-\text{E}(X)^2\) leading to \(\text{E}(X^2)\) |
| \(\text{E}(X^2) = 97\) and \(\text{E}(X) = 9.5\) or \(\text{E}(X^2+X) = 106.5\) or \(3\text{E}(X^2+X) = 319.5\) | A1 | 1.1b |
| \(\text{E}(T) = 3 \times \text{"97"} + 3 \times \text{"9.5"} + 2 \times \frac{68}{15}\) | M1 | 1.1b — method for finding \(\text{E}(T)\) ft their values |
| \(\text{E}(T) = \frac{9857}{30}\) | A1*cso | 2.1 — fully correct solution, no errors; must have \(\text{E}(T) = \frac{9857}{30}\) |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{12}(a-5)^2 = \frac{27}{4}$ | M1 | 3.1a — translating problem into correct equation; allow $\frac{a^3-125}{3(a-5)} - \left(\frac{a+5}{2}\right)^2 = \frac{27}{4}$ |
| $(a-5)^2 = 81$, so $a - 5 = 9$ or $a - 5 = -9$ | A1 | 1.1b — for $a-5=9$ or $a-5=-9$ or equivalent quadratic/cubic |
| Since $a > 5$, $a = 14$ | A1cso* | 2.2a — concluding $a=14$ giving reason why $-4$ is rejected |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct method for $\text{E}(Y)$, $\text{E}(X)$ and $\text{E}(X^2)$ **or** $\text{E}(Y)$ and $\text{E}(X^2+X)$ | M1 | 3.1a — complete method to solve the problem |
| $\text{E}(Y) = \int_2^6 \frac{1}{20} y(2y-3)\, dy$ | M1 | 1.1b — attempt at $\text{E}(Y)$ |
| $= \frac{68}{15}$ | A1 | 1.1b — awrt 4.53 |
| $\text{E}(X) = \frac{5+14}{2} = 9.5$ **and** $\frac{27}{4} = \text{E}(X^2) - 9.5^2$, or $\int_5^{14} \frac{x^2}{9}dx$ or $\int_5^{14}\left(\frac{x^3}{9}+\frac{x^2}{9}\right)dx$ | M1 | 1.1b — attempt at $\text{E}(X)$ and $\text{E}(X^2)$ or $\text{E}(X^2+X)$; allow $\text{Var}(X)=\text{E}(X^2)-\text{E}(X)^2$ leading to $\text{E}(X^2)$ |
| $\text{E}(X^2) = 97$ and $\text{E}(X) = 9.5$ **or** $\text{E}(X^2+X) = 106.5$ **or** $3\text{E}(X^2+X) = 319.5$ | A1 | 1.1b |
| $\text{E}(T) = 3 \times \text{"97"} + 3 \times \text{"9.5"} + 2 \times \frac{68}{15}$ | M1 | 1.1b — method for finding $\text{E}(T)$ ft their values |
| $\text{E}(T) = \frac{9857}{30}$ | A1*cso | 2.1 — fully correct solution, no errors; must have $\text{E}(T) = \frac{9857}{30}$ |\begin{enumerate}
\item The random variable $X$ has a continuous uniform distribution over the interval [5,a], where $a$ is a constant.\\
Given that $\operatorname { Var } ( X ) = \frac { 27 } { 4 }$\\
(a) show that $a = 14$
\end{enumerate}
The continuous random variable $Y$ has probability density function
$$f ( y ) = \left\{ \begin{array} { c c }
\frac { 1 } { 20 } ( 2 y - 3 ) & 2 \leqslant y \leqslant 6 \\
0 & \text { otherwise }
\end{array} \right.$$
The random variable $T = 3 \left( X ^ { 2 } + X \right) + 2 Y$\\
(b) Show that $\mathrm { E } ( T ) = \frac { 9857 } { 30 }$
\hfill \mbox{\textit{Edexcel FS2 AS 2019 Q4 [10]}}