| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Standard +0.3 This is a structured multi-part S3 question requiring recognition of a triangular distribution (symmetric, so E(X)=0 by inspection), routine variance calculation, application of CLT for large sample, and standard confidence interval construction. While it involves several techniques, each step is methodical with no novel insights required—slightly easier than average for Further Maths Statistics. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.03a 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 | Mark | Guidance |
| Graph: two straight line segments forming a tent/triangle shape | G1, G1, G1 | One straight line segment correct; second straight line segment correct; fully labelled intercepts + no spurious other lines. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 0\) (By symmetry) | B1 | |
| \(E(X^2) = \int_{-1}^{0} x^2(1+x)\,dx + \int_{0}^{1} x^2(1-x)\,dx\) | M1 | One correct integral with limits |
| \(= \left[\frac{x^3}{3} + \frac{x^4}{4}\right]_{-1}^{0} + \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}\) | M1 | Second integral correct (with limits) or allow use of symmetry |
| \(= 0 - \left(\frac{-1}{3} + \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) - 0\) | M1 | Correctly integrated and attempt to use limits |
| \(= \frac{1}{6}\), \(\therefore \text{Var}(X) = \frac{1}{6} - 0^2 = \frac{1}{6}\) | A1 | c.a.o. Condone absence of explicit evidence of use of \(\text{Var}(X) = E(X^2) - E(X)^2\). Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{L} \sim N\!\left(k,\, \frac{1}{300}\right)\) | B1 | Normal |
| B1 | Mean | |
| B1 | Variance. ft c's variance in (ii) (\(> 0\)) \(/50\) | |
| Normal distribution because of the Central Limit Theorem | E1 | Any reference to the CLT. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| CI given by \(90.06 \pm 1.96 \times \frac{1}{\sqrt{300}}\) | M1, B1, M1 | |
| \(= 90.06 \pm 0.11316 = (89.947,\ 90.173)\) | A1 | ft c's variance in (ii) (\(>0\)) \(/50\). Must be expressed as an interval. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| It is reasonable, because 90 lies within the interval found in (iv) | E1 | Or equivalent. Total: 1 |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Graph: two straight line segments forming a tent/triangle shape | G1, G1, G1 | One straight line segment correct; second straight line segment correct; fully labelled intercepts + no spurious other lines. Total: 3 |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 0$ (By symmetry) | B1 | |
| $E(X^2) = \int_{-1}^{0} x^2(1+x)\,dx + \int_{0}^{1} x^2(1-x)\,dx$ | M1 | One correct integral with limits |
| $= \left[\frac{x^3}{3} + \frac{x^4}{4}\right]_{-1}^{0} + \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}$ | M1 | Second integral correct (with limits) or allow use of symmetry |
| $= 0 - \left(\frac{-1}{3} + \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) - 0$ | M1 | Correctly integrated and attempt to use limits |
| $= \frac{1}{6}$, $\therefore \text{Var}(X) = \frac{1}{6} - 0^2 = \frac{1}{6}$ | A1 | c.a.o. Condone absence of explicit evidence of use of $\text{Var}(X) = E(X^2) - E(X)^2$. Total: 5 |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{L} \sim N\!\left(k,\, \frac{1}{300}\right)$ | B1 | Normal |
| | B1 | Mean |
| | B1 | Variance. ft c's variance in (ii) ($> 0$) $/50$ |
| Normal distribution because of the Central Limit Theorem | E1 | Any reference to the CLT. Total: 4 |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| CI given by $90.06 \pm 1.96 \times \frac{1}{\sqrt{300}}$ | M1, B1, M1 | |
| $= 90.06 \pm 0.11316 = (89.947,\ 90.173)$ | A1 | ft c's variance in (ii) ($>0$) $/50$. Must be expressed as an interval. Total: 4 |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| It is reasonable, because 90 lies within the interval found in (iv) | E1 | Or equivalent. Total: 1 |4 A timber supplier cuts wooden fence posts from felled trees. The posts are of length $( k + X ) \mathrm { cm }$ where $k$ is a constant and $X$ is a random variable which has probability density function
$$f ( x ) = \begin{cases} 1 + x & - 1 \leqslant x < 0 \\ 1 - x & 0 \leqslant x \leqslant 1 \\ 0 & \text { elsewhere } \end{cases}$$
(i) Sketch $\mathrm { f } ( x )$.\\
(ii) Write down the value of $\mathrm { E } ( X )$ and find $\operatorname { Var } ( X )$.\\
(iii) Write down, in terms of $k$, the approximate distribution of $\bar { L }$, the mean length of a random sample of 50 fence posts. Justify your choice of distribution.\\
(iv) In a particular sample of 50 posts, the mean length is 90.06 cm . Find a $95 \%$ confidence interval for the true mean length of the fence posts.\\
(v) Explain whether it is reasonable to suppose that $k = 90$.
\hfill \mbox{\textit{OCR MEI S3 2011 Q4 [17]}}