Question

Considerapopulationwhoseprobabilitiesaregivenby

p(1)=p(2)=p(3)= 1 3

(a) DetermineE[X]. (b) DetermineSD(X).

σ2σ n =√n

SD(X)=

In the preceding formula, σ is the population standard deviation, and n is the

(c) Let X denote the sample mean of a sample of size 2 from this population. Determine the possible values of X along with their probabilities.

(d) Usetheresultofpart(c)tocomputeE[X]andSD(X).

(e) Areyouranswersconsistent?

Answer 1

a)

x	f(x)	xP(x)	x2P(x)
1.0000	1/3	0.333	0.333
2.0000	1/3	0.667	1.333
3.0000	1/3	1.000	3.000
	total	2.000	4.667
	E(x) =μ=	ΣxP(x) =	2.0000
	E(x2) =	Σx2P(x) =	4.6667
	Var(x)=σ2 =	E(x2)-(E(x))2=	0.6667
	std deviation=	σ= √σ2 =	0.8165

from above E(X)=2

b)SD(X)=0.8165

c)P(X=1)=P(x1=1 ; x2=1) =(1/3)*(!/3) =1/9

P(X=1.5)=P(x1=1 ; x2=2)+P(x1=2 ; x2=1) =(1/3)*(!/3)+(!/3)*(!/3)=2/9

P(X=2)=P(x1=2 ; x2=2)+P(x1=3 ; x2=1)+P(x1=1 ; x2=3) =(1/3)*(!/3)+ (1/3)*(!/3)+(!/3)*(!/3)=3/9=1/3

P(X=2.5)=P(x1=3 ; x2=2)+P(x1=2 ; x2=3) =(1/3)*(!/3)+(!/3)*(!/3)=2/9

P(X=3)=P(x1=3 ; x2=3) =(1/3)*(!/3) =1/9

d)

x	f(x)	xP(x)	x2P(x)
1.0000	1/9	0.111	0.111
1.5000	2/9	0.333	0.500
2.0000	1/3	0.667	1.333
2.5000	2/9	0.556	1.389
3.0000	1/9	0.333	1.000
	total	2.000	4.333
	E(x) =μ=	ΣxP(x) =	2.0000
	E(x2) =	Σx2P(x) =	4.3333
	Var(x)=σ2 =	E(x2)-(E(x))2=	0.3333
	std deviation=	σ= √σ2 =	0.5774

from E(Xbar)=2.000

and SD(Xbar ) =0.5774

e)

as E(Xbar)=E(X)

as well SD(Xbar ) = σ /sqrt(n) =SD(X)/sqrt(2)

therefore answers are consistent