CHAP. 7] PROBABILITY 147
Thus the distributionhofZfollows:
z 371115
h(z) 1 / 62 / 62 / 61 / 6
Hence
E(Z)=∑
zh(z)=
3
6+
14
6+
22
6+
15
6= 9
Accordingly,
E(X+Y)=E(Z)= 9 = 7 + 2 =E(X)+E(Y).BINOMIAL DISTRIBUTION
7.35. The probability that a man hits a target isp= 0 .1. He firesn=100 times. Find the expected numberμ
of times he will hit the target, and the standard deviationσ.
This is a binomial experimentB(n, p)wheren=100,p= 0 .1, andq= 1 −p= 0 .9. Accordingly, we apply
Theorem 7.9 to obtain
μ=np= 100 ( 0. 1 )=10 and σ=
√
np q=√
100 ( 0. 1 )( 0. 9 )= 37.36.A student takes an 18-question multiple-choice exam, with four choices per question. Suppose one of
the choices is obviously incorrect, and the student makes an “educated” guess of the remaining choices.
Find the expected numberE(X)of correct answers, and the standard deviationσ.
This is a biomioal experimentB(n, p)wheren=18,p=^13 , andq= 1 −p=^23. HenceE(X)=np= 18 ·^13 =6 and σ=√
np q=√
18 ·^13 ·^23 = 27.37. The expectation functionE(X)on the space of random variables on a sample spaceScan be proved to be
linear, that is,E(X 1 +X 2 +···+Xn)=E(X 1 )+E(X 2 )+···+E(Xn)Use this property to proveμ=npfor a binomial experimentB(n, p).
On the sample space ofnBernoulli trials, letXi(fori= 1 , 2 ,...,n) be the random variable which has the value 1 or
0 according as theith trial is a success or a failure. Then eachXihas the distribution
x 01
p(x) qp
ThusE(Xi)= 0 (q)+ 1 (p)=p. The total number of successes inntrials is
X=X 1 +X 2 +···+Xn
Using the linearity property ofE, we have
E(X)=E(X 1 +X 2 +···+Xn)
=E(X 1 )+E(X 2 )+···+E(Xn)
=p+p+···+p=npMISCELLANEOUS PROBLEMS
7.38. SupposeXis a random variable with meanμ=75 and standard deviationσ=5.
Estimate the probability thatXlies between 75− 20 =55 and 75+ 20 =95.
Recall Chebyshev’s Inequality statesP(μ−kσ≤X≤μ+kσ)≥ 1 −1
k^2
Herekσ=20. Sinceσ=5, we getk=4. Then, by Chebyshev’s Inequality,P( 55 ≤X≥ 95 )= 1 −
1
42=
15
16≈ 0. 94