MATH 251, Probability and Statistics I, In-class work solutions:
A) X, Y independent. µX
= 3, µY = 4, sigmaX =1, sigmaY
=2 W = 2 - 4X + Y
Mean = 2 -
4·3 + 4= -6
Variance of X = 1, Variance of Y = 22
= 4. Variance of W = (-4)2·1 + 4 = 16 +4
= 20. S.d. of W = sqrt(20)
B) X1, X2, X3
are independent; each has mean 5 and standard deviation 2.
::Find the mean and standard deviation of
their sum, X1 + X2 + X3.
mean = 5+5+5 = 3·5 = 15 = 3·(mean of Xi.)
Variance
of each Xi = 22 = 4. Variance = 4+4+4 =
3·4=12
S.d. = sqrt(12) = sqrt(3·22 ) =
sqrt(3)·2=sqrt(3)·(s.d.
of a single Xi )
::Find the mean and standard deviation of
their average, Xbar = (X1 + X2 + X3)/3
mean = (Mean of (X1 + X2 + X3))/3 =
15/3
= 5 = (mean of Xi.)
Variance
= (variance of (X1 + X2 + X3))·(1/3)2
= 12/9 = ( 3·22 )/(32) = (22
)/3
S.d. = sqrt (12/9) = 2/sqrt(3) = (s.d. of a single Xi
)/sqrt(3)
C) Find the mean and variance and standard
deviation for X:
x 2
3 4 6
p(x) .1 .2
.5 .2
Mean = 2·.1 + 3·.2
+ 4·.5 + 6·.2 = .2 + .6 + 2.0 + 1.2 = 4
Variance =(2-4)2·.1 + (3-4)2·.2
+ (4-4)2·.5 + (6-4)2·.2 =
(-2)2·.1 + (-1)2·.2 +
(0)2·.5
+ (2)2·.2 =
4·.1 + 1·.2
+ 0·.5 + 4·.2 = .4 + .2 + 0 + .8 = 1.4,
s.d. = sqrt(1.4)