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

CHAP. 7] PROBABILITY 135

Binomial Distribution

Consider a binomial experimentB(n, p). That is,B(n, p) consists ofnindependent repeated trials with two
outcomes, success or failure, andpis the probability of success (andq=( 1 −p)is the probability of failure).
The numberXofksuccesses is a random variable with distribution appearing in Fig. 7-5.

Fig. 7-5

The following theorem applies.

Theorem 7.9: Consider the binomial distributionB(n, p). Then:

(i) Expected valueE(X)=μ=np.

(ii) VarianceVar(X)=σ^2 =np q.

(iii) Standard deviationσ=

√

np q.

EXAMPLE 7.17

(a) The probability that a man hits a target isp= 1 /5. He fires 100 times. Find the expected numberμof times

he will hit the target and the standard deviationσ.

Herep=^15 and soq=^45. Hence

μ=np= 100 ·

1

5

=20 and σ=

√

np q=

√

100 ·

1

5

·

4

5

= 4

(b) Find the expected numberE(X)of correct answers obtained by guessing in a five-question true–false test.

Herep=^12. HenceE(X)=np= 5 ·^12 = 2 .5.

7.8 CHEBYSHEV’S INEQUALITY, LAW OF LARGE NUMBERS

The standard deviationσof a random variableXmeasures the weighted spread of the values ofXabout
the meanμ. Thus, for smallerσ, we would expect thatXwill be closer toμ. A more precise statement of
this expectation is given by the following inequality, named after the Russian mathematician P. L. Chebyshev
(1821–1894).

Theorem 7.10 (Chebyshev’s Inequality): LetXbe a random variable with meanμand standard deviationσ.
Thenforanypositivenumberk, theprobabilitythatavalueofXliesintheinterval[μ−kσ, μ+kσ]
is at least 1− 1 /k^2. That is,

P(μ−kσ≤X≤μ+kσ)≥ 1 −

1

k^2

EXAMPLE 7.18 SupposeXis a random variable with meanμ=75 and standard deviationσ =5. What
conclusion aboutXcan be drawn from Chebyshev’s inequality fork=2 andk=3?
Settingk=2, we obtain:

μ−kσ= 75 − 2 ( 5 )=65 and μ+kσ= 75 + 2 ( 5 )= 85