| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Estimated variance confidence interval |
| Difficulty | Challenging +1.2 This is a multi-part question combining continuous distributions with confidence intervals. Parts (i)-(iii) involve standard pdf techniques (symmetry argument, integration for k, variance calculation). Part (iv) requires applying the CLT to construct a confidence interval, which is routine S3 material. Part (v) tests conceptual understanding of confidence intervals. While it requires multiple techniques and careful working, each component is a standard textbook exercise with no novel insight required, making it moderately above average difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Triangular shape with peak at \(ka\) and base ending at \(2a\) | G1 | Shape |
| Correct scales on axes shown | G1 | Scales on axes |
| \(E(X) = a\) because the distribution is symmetrical about \(x = a\) | B1 | Do not allow integration method |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total area \(= \frac{1}{2} \cdot 2a \cdot ka\) or \(\int_0^a kx\,dx + \int_a^{2a} k(2a-x)\,dx\) | M1 | Attempting to find area of triangle or setting up correct integrals including limits |
| \(k\left[\frac{x^2}{2}\right]_0^a + k\left[2ax - \frac{x^2}{2}\right]_a^{2a}\) leading to \(k\left(\frac{a^2}{2}-0\right)+k\left(2a^2-\frac{3a^2}{2}\right) = ka^2\) | A1 | Correctly finding area in terms of \(k, a\) |
| \(ka^2 = 1 \Rightarrow k = \frac{1}{a^2}\) | A1 | Equating area to 1 and convincingly obtaining result. Answer given |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}\,X = k\int_0^a x^3\,dx + k\int_a^{2a}(2ax^2 - x^3)\,dx - a^2\) | M1 | Correct integral for \(E(X^2)\) including limits |
| \(k\left[\frac{x^4}{4}\right]_0^a + k\left[\frac{2ax^3}{3} - \frac{x^4}{4}\right]_a^{2a} - a^2\) | M1 | Correctly integrated (dependent on M1 above) |
| \(\frac{a^2}{4} + \frac{16a^2}{3} - 4a^2 - \frac{2a^2}{3} + \frac{a^2}{4} - a^2\) | M1 | Using \(E(X^2) - (E(X))^2\) |
| \(\text{Var}\,X = \frac{a^2}{6}\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 1.92\), \(s = 0.8352\) | B1 | 1.96 seen |
| Interval is \(1.92 \pm 1.96\dfrac{0.8352}{\sqrt{50}}\) | M1 | Correct SE |
| M1 | Centred on 1.92 | |
| \(= (1.69,\ 2.15)\) | A1 | cao |
| The distribution of \(\bar{X}\) is approximately Normal (CLT) or \(s\) is only an estimate | E1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| This statement is incorrect | E1 | |
| The value of \(a\) either lies within this particular interval or it does not — so the probability is either 0 or 1. If a large number of such intervals were constructed, then \(a\) would lie in 95% of them. | E1 | A comment either about \(p=0\) or \(1\), or about a large number of intervals |
| [2] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Triangular shape with peak at $ka$ and base ending at $2a$ | G1 | Shape |
| Correct scales on axes shown | G1 | Scales on axes |
| $E(X) = a$ because the distribution is symmetrical about $x = a$ | B1 | Do not allow integration method |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total area $= \frac{1}{2} \cdot 2a \cdot ka$ or $\int_0^a kx\,dx + \int_a^{2a} k(2a-x)\,dx$ | M1 | Attempting to find area of triangle or setting up correct integrals including limits |
| $k\left[\frac{x^2}{2}\right]_0^a + k\left[2ax - \frac{x^2}{2}\right]_a^{2a}$ leading to $k\left(\frac{a^2}{2}-0\right)+k\left(2a^2-\frac{3a^2}{2}\right) = ka^2$ | A1 | Correctly finding area in terms of $k, a$ |
| $ka^2 = 1 \Rightarrow k = \frac{1}{a^2}$ | A1 | Equating area to 1 and convincingly obtaining result. Answer given |
| **[3]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}\,X = k\int_0^a x^3\,dx + k\int_a^{2a}(2ax^2 - x^3)\,dx - a^2$ | M1 | Correct integral for $E(X^2)$ including limits |
| $k\left[\frac{x^4}{4}\right]_0^a + k\left[\frac{2ax^3}{3} - \frac{x^4}{4}\right]_a^{2a} - a^2$ | M1 | Correctly integrated (dependent on M1 above) |
| $\frac{a^2}{4} + \frac{16a^2}{3} - 4a^2 - \frac{2a^2}{3} + \frac{a^2}{4} - a^2$ | M1 | Using $E(X^2) - (E(X))^2$ |
| $\text{Var}\,X = \frac{a^2}{6}$ | A1 | cao |
| **[4]** | | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 1.92$, $s = 0.8352$ | B1 | 1.96 seen |
| Interval is $1.92 \pm 1.96\dfrac{0.8352}{\sqrt{50}}$ | M1 | Correct SE |
| | M1 | Centred on 1.92 |
| $= (1.69,\ 2.15)$ | A1 | cao |
| The distribution of $\bar{X}$ is approximately Normal (CLT) or $s$ is only an estimate | E1 | |
| **[5]** | | |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| This statement is incorrect | E1 | |
| The value of $a$ either lies within this particular interval or it does not — so the probability is either 0 or 1. If a large number of such intervals were constructed, then $a$ would lie in 95% of them. | E1 | A comment either about $p=0$ or $1$, or about a large number of intervals |
| **[2]** | | |4 The probability density function of a random variable $X$ is given by
$$\mathrm { f } ( x ) = \begin{cases} k x & 0 \leqslant x \leqslant a \\ k ( 2 a - x ) & a < x \leqslant 2 a \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $k$ are positive constants.\\
(i) Sketch $\mathrm { f } ( x )$. Hence explain why $\mathrm { E } ( X ) = a$.\\
(ii) Show that $k = \frac { 1 } { a ^ { 2 } }$.\\
(iii) Find $\operatorname { Var } ( X )$ in terms of $a$.
In order to estimate the value of $a$, a random sample of size 50 is taken from the distribution. It is found that the sample mean and standard deviation are $\bar { x } = 1.92$ and $s = 0.8352$.\\
(iv) Construct a symmetrical $95 \%$ confidence interval for $a$. Give one reason why the answer is only approximate.\\
(v) A non-statistician states that the probability that $a$ lies in the interval found in part (iv) is 0.95 . Comment on this statement.
\section*{END OF QUESTION PAPER}
\section*{OCR $^ { \text {® } }$}
\hfill \mbox{\textit{OCR MEI S3 2014 Q4 [17]}}