| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine application of continuous distributions, binomial approximation by normal, and CLT. Part (i) requires variance calculation from a pdf (straightforward integration), part (ii) is direct probability calculation, part (iii) applies normal approximation to binomial (standard technique), and part (iv) simply requires stating the CLT result. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2}{9}\int_0^3 x^3(3-x)\,dx = \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3 = 2.7\) | M1, A1 | Integrate \(x^2 f(x)\) from 0 to 3 [*not* for \(\mu\)]. Correct indefinite integral |
| Mean is \(1\frac{1}{2}\), soi | B1 | [not recoverable later] |
| \((1\tfrac{1}{2})^2 = \frac{9}{20}\) or \(\mathbf{0.45}\) | M1, A1 5 | Subtract their \(\mu^2\). Answer art 0.450 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2}{9}\int_0^{0.5} x(3-x)\,dx = \frac{2}{9}\left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^{0.5}\) | M1 | Integrate \(f(x)\) between 0, 0.5, must be seen somewhere |
| \(= \frac{2}{27}\) AG | A1 2 | Correctly obtain given answer \(\frac{2}{27}\), decimals other than 0.5 not allowed, 1 more line needed (eg \([\ ] = \frac{1}{3}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B\!\left(108,\ \frac{2}{27}\right)\) | B1 | \(B(108,\ \frac{2}{27})\) seen or implied, eg \(Po(8)\) |
| \(\approx N(8,\ 7.4074)\) | M1, A1 | Normal, mean 8 … variance (or SD) \(200/27\) or art 7.41 |
| \(1 - \Phi\!\left(\dfrac{9.5 - 8}{\sqrt{7.4074}}\right)\) | M1 | Standardise 10, allow \(\sqrt{}\) errors, wrong or no cc, needs to be using \(B(108,\ldots)\) |
| \(= 1 - \Phi(0.5511) = \mathbf{0.291}\) | A1, A1 6 | Correct \(\sqrt{}\) and cc. Final answer, art 0.291 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{X} \sim N\!\left(1.5,\ \frac{1}{240}\right)\) | B1, B1\(\sqrt{}\), B1\(\sqrt{}\) 3 | Normal. NB: *not* part (iii). Mean their \(\mu\). Variance or SD: (their \(0.45\))/108 [not \((8,\ 50/729)\)] |
# Question 7:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2}{9}\int_0^3 x^3(3-x)\,dx = \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3 = 2.7$ | M1, A1 | Integrate $x^2 f(x)$ from 0 to 3 [*not* for $\mu$]. Correct indefinite integral |
| Mean is $1\frac{1}{2}$, soi | B1 | [not recoverable later] |
| $(1\tfrac{1}{2})^2 = \frac{9}{20}$ or $\mathbf{0.45}$ | M1, A1 **5** | Subtract their $\mu^2$. Answer art 0.450 |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2}{9}\int_0^{0.5} x(3-x)\,dx = \frac{2}{9}\left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^{0.5}$ | M1 | Integrate $f(x)$ between 0, 0.5, must be seen somewhere |
| $= \frac{2}{27}$ AG | A1 **2** | Correctly obtain given answer $\frac{2}{27}$, decimals other than 0.5 not allowed, 1 more line needed (eg $[\ ] = \frac{1}{3}$) |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $B\!\left(108,\ \frac{2}{27}\right)$ | B1 | $B(108,\ \frac{2}{27})$ seen or implied, eg $Po(8)$ |
| $\approx N(8,\ 7.4074)$ | M1, A1 | Normal, mean 8 … variance (or SD) $200/27$ or art 7.41 |
| $1 - \Phi\!\left(\dfrac{9.5 - 8}{\sqrt{7.4074}}\right)$ | M1 | Standardise 10, allow $\sqrt{}$ errors, wrong or no cc, needs to be using $B(108,\ldots)$ |
| $= 1 - \Phi(0.5511) = \mathbf{0.291}$ | A1, A1 **6** | Correct $\sqrt{}$ and cc. Final answer, art 0.291 |
---
# Question 7 (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{X} \sim N\!\left(1.5,\ \frac{1}{240}\right)$ | B1, B1$\sqrt{}$, B1$\sqrt{}$ **3** | Normal. NB: *not* part (iii). Mean their $\mu$. Variance or SD: (their $0.45$)/108 [not $(8,\ 50/729)$] |
---7 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 2 } { 9 } x ( 3 - x ) & 0 \leqslant x \leqslant 3 , \\ 0 & \text { otherwise } . \end{cases}$$
(i) Find the variance of $X$.\\
(ii) Show that the probability that a single observation of $X$ lies between 0.0 and 0.5 is $\frac { 2 } { 27 }$.\\
(iii) 108 observations of $X$ are obtained. Using a suitable approximation, find the probability that at least 10 of the observations lie between 0.0 and 0.5 .\\
(iv) The mean of 108 observations of $X$ is denoted by $\bar { X }$. Write down the approximate distribution of $\bar { X }$, giving the value(s) of any parameter(s).
\hfill \mbox{\textit{OCR S2 2009 Q7 [16]}}