Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

(Martin Jones) #1

134 PROBABILITY [CHAP. 7


EXAMPLE 7.15


(a) Suppose a fair coin is tossed six times. The number of heads which can occur with their respective
probabilities follows:

xi 0123456
p 1 1 /64 6/64 15/64 20/64 15/64 6/64 1/ 64

Then the mean or expectation (or expected number of heads) is:

μ=E(X)= 0

(
1
64

)
+ 1

(
6
64

)
+ 2

(
15
64

)
+ 3

(
20
64

)
+ 4

(
15
64

)
+ 5

(
6
64

)
+ 6

(
1
64

)
= 3

(This agrees with our intuition that we expect that half of the tosses to be heads.)

(b) Three horsesa,b, andcare in a race; suppose their respective probabilities of winning are^12 ,^13 , and^16. Let
Xdenote the payoff function for the winning horse, and supposeXpays $2, $6, or $9 according asa,b,or
cwins the race. The expected payoff for the race is


E(X)=X(a)P (a)+X(b)P (b)+X(c)P (c)

= 2

(
1
2

)
+ 6

(
1
3

)
+ 9

(
1
6

)
= 4. 5

Variance and Standard Deviation of a Random Variable


LetXbe a random variable with meanμand distributionfas in Fig. 7-4. Then thevarianceofX, denoted
byVar(X), is defined by:


Var(X)=(x 1 −μ)^2 f(x 1 )+(x 2 −μ)^2 f(x 2 )+···+(xt−μ)^2 f(xt)=(xk−μ)^2 f(xk)=E((X−μ)^2 )

Alternately, when the notation[xk,pk]is used instead of[xk,f(xk)],


Var(X)=(x 1 −μ)^2 p 1 +(x 2 −μ)^2 p 2 +···+(xt−μ)^2 pt=(xk−μ)^2 pk=E((X−μ)^2 )

Thestandard deviationofX, denoted byσxor simplyσ, is the nonnegative square root ofVar(X):


σx=


Var(X)

Accordingly,Var(X)=σx^2. BothVar(X)andσx^2 or simplyσ^2 are used to denote the variance ofX.
The following formulas are usually more convenient for computingVar(X)than the above:


Var(X)=x 12 f(x 1 )+x 22 f(x 2 )+···+xt^2 f(xt)−μ^2 =

[∑
xk^2 f(xk)

]
−μ^2 =E(X^2 )−μ^2

or


Var(X)=x 12 p 1 +x 22 p 2 +···+xt^2 pt−μ^2 =

[∑
x^2 kpk

]
−μ^2 =E(X^2 )−μ^2

EXAMPLE 7.16 LetXdenote the number of times heads occurs when a fair coin is tossed six times. The
distribution ofXappears in Example 7.15(a), where its meanμ=3 is computed. The variance ofXis computed
as follows:
Var(X)=( 0 − 3 )^2641 +( 1 − 3 )^2646 +( 2 − 3 )^21564 +···+( 6 − 3 ) 641 = 1. 5


Alternatively:


Var(X)= 02641 + 12646 + 221564 + 322064 + 421564 + 52646 + 62641 − 32 = 1. 5

Thus the standard deviation isσ=



1. 5 ≈ 1 .225 (heads).
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