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Considerapopulationwhoseprobabilitiesaregivenby p(1)=p(2)=p(3)= 1 3 (a) DetermineE[X]. (b) DetermineSD(X). ?2? n =?n SD(X)= In the preceding formula, ? is the population...

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Basic Math

Considerapopulationwhoseprobabilitiesaregivenby
p(1)=p(2)=p(3)= 1 3
(a) DetermineE[X]. (b) DetermineSD(X).
?2? n =?n
SD(X)=
In the preceding formula, ? is the population standarddeviation, and n is the
(c) Let X denote the sample mean of a sample of size 2 fromthis population. Determine the possible values of X along withtheir probabilities.
(d) Usetheresultofpart(c)tocomputeE[X]andSD(X).
(e) Areyouranswersconsistent?

Answer & Explanation Solved by verified expert
4.0 Ratings (670 Votes)

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) =?= ?xP(x) = 2.0000
E(x2) = ?x2P(x) = 4.6667
Var(x)=?2 = E(x2)-(E(x))2= 0.6667
std deviation=         ?= ??2 = 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) =?= ?xP(x) = 2.0000
E(x2) = ?x2P(x) = 4.3333
Var(x)=?2 = E(x2)-(E(x))2= 0.3333
std deviation=         ?= ??2 = 0.5774

from E(Xbar)=2.000

and SD(Xbar ) =0.5774

e)

as E(Xbar)=E(X)

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

therefore answers are consistent
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