Date | May 2014 | Marks available | 5 | Reference code | 14M.3sp.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
The random variable X has probability distribution Po(8).
(i) Find \({\text{P}}(X = 6)\).
(ii) Find \({\text{P}}(X = 6|5 \leqslant X \leqslant 8)\).
\(\bar X\) denotes the sample mean of \(n > 1\) independent observations from \(X\).
(i) Write down \({\text{E}}(\bar X)\) and \({\text{Var}}(\bar X)\).
(ii) Hence, give a reason why \(\bar X\) is not a Poisson distribution.
A random sample of \(40\) observations is taken from the distribution for \(X\).
(i) Find \({\text{P}}(7.1 < \bar X < 8.5)\).
(ii) Given that \({\text{P}}\left( {\left| {\bar X - 8} \right| \leqslant k} \right) = 0.95\), find the value of \(k\).
Markscheme
(i) \({\text{P}}(X = 6) = 0.122\) (M1)A1
(ii) \({\text{P}}(X = 6|5 \leqslant X \leqslant 8) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(5 \leqslant X \leqslant 8)}} = \frac{{0.122 \ldots }}{{0.592 \ldots - 0.0996 \ldots }}\) (M1)(A1)
\( = 0.248\) A1
[5 marks]
(i) \({\text{E}}(\bar X) = 8\) A1
\({\text{Var}}(\bar X) = \frac{8}{n}\) A1
(ii) \({\text{E}}(\bar X) \ne {\text{Var}}(\bar X)\) \({\text{(for }}n > 1)\) R1
Note: Only award the R1 if the two expressions in (b)(i) are different.
[3 marks]
(i) EITHER
\(\bar X \sim {\text{N(8, 0.2)}}\) (M1)A1
Note: M1 for normality, A1 for parameters.
\({\text{P}}(7.1 < \bar X < 8.5) = 0.846\) A1
OR
The expression is equivalent to
\({\text{P}}(283 \leqslant \sum {X \leqslant 339)} \) where \(\sum X \) is \({\text{Po(320)}}\) M1A1
\( = 0.840\) A1
Note: Accept 284, 340 instead of 283, 339
Accept any answer that rounds correctly to 0.84 or 0.85.
(ii) EITHER
\(k = 1.96\frac{\sigma }{{\sqrt n }}\) or \(1.96{\text{ std}}(\bar X)\) (M1)(A1)
\(k = 0.877\) or \(1.96\sqrt {0.2} \) A1
OR
The expression is equivalent to
\(P(320 - 40k \leqslant \sum {X \leqslant 320 + 40k) = 0.95} \) (M1)
\(k = 0.875\) A2
Note: Accept any answer that rounds to 0.87 or 0.88.
Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75
[6 marks]