| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Moderate -0.3 This is a straightforward application of standard uniform distribution formulas. Part (a) requires recognizing that the pdf integrates to 1, giving k = b - a. Part (b)(i) involves substituting the known formulas for mean and variance of a uniform distribution into two equations and solving simultaneously. Part (b)(ii) is a simple binomial probability calculation once the interval is found. All steps are routine with no novel 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = b - a\) | B1 | Must be in terms of \(a\) and \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{a+b}{2} = 1\) so \(a + b = 2\) | B1 | Correct mean equation |
| \(Var(X) = \frac{(b-a)^2}{12} = 3\) so \((b-a)^2 = 36\), \(b - a = 6\) | M1 A1 | M1 for correct variance formula used, A1 for correct equation |
| \(a = -2\), \(b = 4\) | A1 | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 0) = \frac{0-(-2)}{4-(-2)} = \frac{2}{6} = \frac{1}{3}\) | B1 | Correct probability of negative value |
| \(P(\text{exactly one negative}) = \binom{4}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^3\) | M1 | Correct binomial structure |
| \(= \frac{32}{81} \approx 0.395\) | A1 | Correct final answer |
# Question 2:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = b - a$ | B1 | Must be in terms of $a$ and $b$ |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{a+b}{2} = 1$ so $a + b = 2$ | B1 | Correct mean equation |
| $Var(X) = \frac{(b-a)^2}{12} = 3$ so $(b-a)^2 = 36$, $b - a = 6$ | M1 A1 | M1 for correct variance formula used, A1 for correct equation |
| $a = -2$, $b = 4$ | A1 | Both values correct |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 0) = \frac{0-(-2)}{4-(-2)} = \frac{2}{6} = \frac{1}{3}$ | B1 | Correct probability of negative value |
| $P(\text{exactly one negative}) = \binom{4}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^3$ | M1 | Correct binomial structure |
| $= \frac{32}{81} \approx 0.395$ | A1 | Correct final answer |
---2 The continuous random variable $X$ has probability density function defined by
$$f ( x ) = \begin{cases} \frac { 1 } { k } & a \leqslant x \leqslant b \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Write down, in terms of $a$ and $b$, the value of $k$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { E } ( X ) = 1$ and $\operatorname { Var } ( X ) = 3$, find the values of $a$ and $b$.
\item Four independent values of $X$ are taken. Find the probability that exactly one of these four values is negative.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2015 Q2 [8]}}