| Date | May 2013 | Marks available | 5 | Reference code | 13M.1.hl.TZ0.5 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Let \({X_k}\) be independent normal random variables, where \({\rm{E}}({X_k}) = \mu \) and \(Var({X_k}) = \sqrt k \) , for \(k = 1,2, \ldots \) .
The random variable \(Y\) is defined by \(Y = \sum\limits_{k = 1}^6 {\frac{{{{( - 1)}^{k + 1}}}}{{\sqrt k }}} {X_k}\) .
(i) Find \({\rm{E}}(Y)\) in the form \(p\mu \) , where \(p \in \mathbb{R}\) .
(ii) Find \(k\) if \({\rm{Var}}({X_k}) < {\rm{Var}}(Y) < {\rm{Var}}({X_{k + 1}})\) .
A random sample of \(n\) values of \(Y\) was found to have a mean of \(8.76\).
(i) Given that \(n = 10\) , determine a \(95\%\) confidence interval for \(\mu \) .
(ii) The width of the confidence interval needs to be halved. Find the appropriate value of \(n\) .
Markscheme
(i) \({\rm{E}}(Y) = \frac{1}{{\sqrt 1 }}\mu - \frac{1}{{\sqrt 2 }}\mu + \frac{1}{{\sqrt 3 }}\mu - \frac{1}{{\sqrt 4 }}\mu + \frac{1}{{\sqrt 5 }}\mu - \frac{1}{{\sqrt 6 }}\mu \) (M1)
\( = 0.409\) \((209)\mu \) A1
Note: Accept answers which round to \(0.41\).
(ii) \(Var(Y) = \frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 4 }} + \frac{1}{{\sqrt 5 }} + \frac{1}{{\sqrt 6 }}\) (M1)
\( = 3.64\) \((3.6399 \ldots )\) A1
\(({\rm{Var}}({X_{13}}) = 3.61;{\rm{Var}}({X_{14}}) = 3.74) \Rightarrow k = 13\) A1
[5 marks]
(i) \(95\%\) CI for is \({\rm{E}}(Y)\)
\(8.76 \pm 1.96\sqrt {\frac{{3.6399 \ldots }}{{10}}} \) (M1)
\( = \left[ {7.58,9.94} \right]\) A1A1
Note: Accept \(\left[ {7.6,9.9} \right]\) . Do not penalize answers given to more than 3sf.
Since \(\mu = \frac{{{\rm{E}}(Y)}}{{0.409 \ldots }}\) , CI for \(\mu \) is \(\left[ {18.5,24.3} \right]\) A1
Note: Do not penalize answers given to more than 3sf.
(ii) width of a CI is inversely proportional to the square root of \(n\) (M1)
so \(n = 40\) A1
[6 marks]
Examiners report
This question proved to be the most difficult on the paper and few fully correct answers were seen. In part (a) (i) many candidates did know how to find the answers in terms of \(\mu\). Very few candidates successfully completed part (a) (ii).
This question proved to be the most difficult on the paper and few fully correct answers were seen. In part b) (i) a number of candidates made some progress, but few realised or knew how to convert the confidence interval for \({\rm{E}}(Y)\ into a confidence interval for \(\mu\). For those who persevered to the end of the question, there was a reasonable degree of success in part (b) (ii).
