| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 11 |
| Topic | Linear combinations of normal random variables |
| Type | Estimated variance confidence interval |
| Difficulty | Standard +0.3 This is a straightforward application of standard Further Statistics techniques: finding unbiased estimates from sample data, working backwards to find Σc², applying the normal distribution property for sums, and combining independent normal variables. All steps are routine calculations with no novel insight required, though it does test understanding of multiple concepts across sampling theory and distribution of linear combinations. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = \bar{c} = \frac{224.0}{16} = 14\) | B1 [1] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{16\left(\sum c^2 - \bar{c}^2\right)}{15} \left(\frac{\sum c^2}{16} - \bar{c}^2\right) = 0.24\) | M1 | 2.1 |
| M1 | 3.1a | 16/15 factor used |
| \(\left(\frac{\sum c^2 - 14^2}{16}\right) = 0.225\) | ||
| \(\sum c^2 = 3139.6\) | A1 | 1.1 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Yes as the sum of independent normal RVs is normal | B1 [1] | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(D) = 10 \times 0.24 = 2.4\) | B1 | 1.1 |
| \(\text{SD}(F) = (13.2 - 15.0) \div \Phi^{-1}(0.115)\) | M1 | 1.1a |
| \(= 1.50\) | A1 | 1.1 |
| \(D + F \sim N(153.0, 4.65)\) | M1 | 3.1a |
| \(P(> 155.0) = 1 - 0.82316\ldots\) | M1 | 1.1a |
| \(= 0.176839\ldots\) | A1 [6] | 1.1 |
## (i)
$\mu = \bar{c} = \frac{224.0}{16} = 14$ | B1 [1] | 1.1 | 14 or 14.0 only
## (ii)
$\frac{16\left(\sum c^2 - \bar{c}^2\right)}{15} \left(\frac{\sum c^2}{16} - \bar{c}^2\right) = 0.24$ | M1 | 2.1 | Correct formula used | Allow if no 16/15
| M1 | 3.1a | 16/15 factor used
$\left(\frac{\sum c^2 - 14^2}{16}\right) = 0.225$ | | |
$\sum c^2 = 3139.6$ | A1 | 1.1 | 3139.6 or exact equivalent only
| [3] |
## (iii)
Yes as the sum of independent normal RVs is normal | B1 [1] | 2.4 | Any mention of CLT: B0
## (iv)
$\text{Var}(D) = 10 \times 0.24 = 2.4$ | B1 | 1.1 | Variance 2.4
$\text{SD}(F) = (13.2 - 15.0) \div \Phi^{-1}(0.115)$ | M1 | 1.1a | Use $\Phi^{-1}$ to find SD of $F$ | Allow SD/var muddle for M1
$= 1.50$ | A1 | 1.1 |
$D + F \sim N(153.0, 4.65)$ | M1 | 3.1a | Combine, add variances | Allow if SDs used
$P(> 155.0) = 1 - 0.82316\ldots$ | M1 | 1.1a | 0.348 from $\sigma^2 = 1.5^2 + 24$: M1A0
$= 0.176839\ldots$ | A1 [6] | 1.1 | Awrt 0.1779 The continuous random variable $C$ has the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. The sum of a random sample of 16 observations of $C$ is 224.0 .\\
(i) Find an unbiased estimate of $\mu$.\\
(ii) It is given that an unbiased estimate of $\sigma ^ { 2 }$ is 0.24. Find the value of $\Sigma c ^ { 2 }$.\\
$D$ is the sum of 10 independent observations of $C$.\\
(iii) Explain whether $D$ has a normal distribution.
The continuous random variable $F$ is normally distributed with mean 15.0, and it is known that $\mathrm { P } ( F < 13.2 ) = 0.115$.\\
(iv) Use the unbiased estimates of $\mu$ and $\sigma ^ { 2 }$ to find $\mathrm { P } ( D + F > 157.0 )$.
\section*{OCR}
\section*{Oxford Cambridge and RSA}
\hfill \mbox{\textit{OCR Further Statistics 2018 Q9 [11]}}