| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Conditional probability with uniform |
| Difficulty | Moderate -0.8 This is a straightforward Further Statistics question testing basic understanding of uniform distributions and linear transformations. Parts (a)-(c) are routine recall and simple calculations, while part (d) involves conditional probability but with a uniform distribution where the calculation is mechanical (finding P(X>2.5 and X<7.5)/P(X<7.5)). No novel insight or complex problem-solving required. |
| Spec | 5.02e Discrete uniform distribution5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| VILU SIHI NI IIIUM ION OC | VGHV SIHILNI IMAM ION OO | VJYV SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| (Continuous) uniform or rectangular | B1 | Discrete uniform is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([P(Y>0) = P(5-2X>0) =] P(X < 2.5)\) or \(f(y) = \left\{\frac{1}{16},\quad -13 \leq y \leq 3\right\}\) | M1 | Using distribution of \(X\) to obtain \(P(X<2.5)\) or finding distribution of \(Y\) in range \(-13 \leq y \leq 3\) |
| \(\frac{2.5-1}{8} = \frac{3}{16}\) or \(\frac{3-0}{16} = \frac{3}{16}\) | A1 | \(\frac{3}{16}\) or 0.1875 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(Y) = 5 - 2E(X)\), \([= 5 - 2(\frac{1+9}{2})]\) or \(E(Y) = \frac{-13+3}{2}\) | M1 | Use of \(E(aX+b)\) or use of \(\frac{a+b}{2}\) from distribution of \(Y\) |
| \(= \mathbf{-5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([P(Y<0) \mid (X<7.5)] = \frac{P(2.5 < X < 7.5)}{P(X < 7.5)}\) | M1 | Correct ratio expression |
| \(= \frac{\frac{7.5-2.5}{8}}{\frac{7.5-1}{8}} \left[= \frac{0.625}{0.8125}\right]\) | M1 | Correct numerical expression |
| \(= \frac{10}{13}\) | A1 | SC: If M0M0A0, a correct numerator or correct denominator scores M0M1A0 |
# Question 2:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| (Continuous) uniform or rectangular | B1 | Discrete uniform is B0 |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $[P(Y>0) = P(5-2X>0) =] P(X < 2.5)$ or $f(y) = \left\{\frac{1}{16},\quad -13 \leq y \leq 3\right\}$ | M1 | Using distribution of $X$ to obtain $P(X<2.5)$ or finding distribution of $Y$ in range $-13 \leq y \leq 3$ |
| $\frac{2.5-1}{8} = \frac{3}{16}$ or $\frac{3-0}{16} = \frac{3}{16}$ | A1 | $\frac{3}{16}$ or 0.1875 |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $E(Y) = 5 - 2E(X)$, $[= 5 - 2(\frac{1+9}{2})]$ or $E(Y) = \frac{-13+3}{2}$ | M1 | Use of $E(aX+b)$ or use of $\frac{a+b}{2}$ from distribution of $Y$ |
| $= \mathbf{-5}$ | A1 | |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| $[P(Y<0) \mid (X<7.5)] = \frac{P(2.5 < X < 7.5)}{P(X < 7.5)}$ | M1 | Correct ratio expression |
| $= \frac{\frac{7.5-2.5}{8}}{\frac{7.5-1}{8}} \left[= \frac{0.625}{0.8125}\right]$ | M1 | Correct numerical expression |
| $= \frac{10}{13}$ | A1 | SC: If M0M0A0, a correct numerator or correct denominator scores M0M1A0 |\begin{enumerate}
\item The continuous random variable X has probability density function
\end{enumerate}
$$f ( x ) = \begin{cases} \frac { 1 } { 8 } & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$
(a) Write down the name given to this distribution.
The continuous random variable $Y = 5 - 2 X$\\
(b) Find $\mathrm { P } ( Y > 0 )$\\
(c) Find $\mathrm { E } ( Y )$\\
(d) Find $\mathrm { P } ( Y < 0 \mid X < 7.5 )$
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VILU SIHI NI IIIUM ION OC & VGHV SIHILNI IMAM ION OO & VJYV SIHI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel FS2 AS 2018 Q2 [8]}}