ENGR 323
HE, D.
3-40
BEAUTIFUL HOMEWORK #4
3-40
1/1
Continuation of Exercise 3-16. Determine the mean and variance of the random variable
in Exercise 3-16.
Recall the solution from Exercise 3-16, the discrete random variable X denotes the
yearly revenue (millions of dollars) of the product, the range of RV is X = {1, 5, 10},
and for all x ∈ X, their probabilities are;
fX(1) = P(X = 1) = 0.1
fX(5) = P(X = 5) = 0.6
fX(10) = P(X = 10) = 0.3
Definition of mean or expected value of the discrete random variable X is on
page 113 of textbook, it denoted as µX or E(X), is
µX = E(X) =
∑ xf
X
(x)
x
So, the mean or expected value for cab be calculate using the above equation,
µX = E(X) =
∑ xf
X
(x)
x
= (1)*fX(1) + (5)*fX(5) + (10)*fX(10)
= 1(0.1) + 5(0.6) + 10(0.3)
= 6.1 (millions of dollars)
Definition of variance of a random variable X is on page 116 of textbook, it
denotes as σ X2 or V(X), is
σ X2 = V(X) = E(X - µX)2 = ∑ (x − µX ) 2 fX (x)
x
So, the variance of the discrete RV X can be determine using the above
equation,
σ X2 = V(X) = E(X - µX)2 = ∑ (x − µX ) 2 fX (x)
x
= (1- 6.1)2(0.1) + (5 - 6.1)2(0.6) + (10 - 6.1)2(0.3)
= 7.89 (millions of dollars)2
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