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⩽X⩽8).
ˉX denotes the sample mean of n>1 independent observations from X.
(i) Write down E(ˉX) and Var(ˉX).
(ii) Hence, give a reason why ˉX is not a Poisson distribution.
A random sample of 40 observations is taken from the distribution for X.
(i) Find P(7.1<ˉX<8.5).
(ii) Given that P(|ˉX−8|⩽k)=0.95, find the value of k.
Markscheme
(i) P(X=6)=0.122 (M1)A1
(ii) P(X=6|5⩽X⩽8)=P(X=6)P(5⩽X⩽8)=0.122…0.592…−0.0996… (M1)(A1)
=0.248 A1
[5 marks]
(i) E(ˉX)=8 A1
Var(ˉX)=8n A1
(ii) E(ˉX)≠Var(ˉX) (for n>1) R1
Note: Only award the R1 if the two expressions in (b)(i) are different.
[3 marks]
(i) EITHER
ˉX∼N(8, 0.2) (M1)A1
Note: M1 for normality, A1 for parameters.
P(7.1<ˉX<8.5)=0.846 A1
OR
The expression is equivalent to
P(283⩽∑X⩽339) where ∑X is 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=1.96σ√n or 1.96 std(ˉX) (M1)(A1)
k=0.877 or 1.96√0.2 A1
OR
The expression is equivalent to
P(320−40k⩽∑X⩽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]