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 P(X=6).
(ii) Find P(X=6|5⩽.
\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]