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

148 PROBABILITY [CHAP. 7

7.39. LetXbe a random variable with meanμ=40 and standard deviationσ=2. Use Chebyshev’s Inequality

to find abfor whichP( 40 −b≤X≤ 40 +b)≥ 0 .95.

First solve 1− 1 /k^2 = 0 .95 forkas follows:

0. 05 =

1

k^2

or k^2 =

1

0. 05

=20 or k=

√

20 = 2

√

5

Then, by Chebyshev’s Inequality,b=kσ= 10

√

5 ≈ 23 .4. Hence[P( 16. 6 ≤X≤ 63. 60 )≥ 0. 95 ]

7.40. LetXbe a random variable with distributionf. TherthmomentMrofXis defined by

Mr=E(Xr)=

∑

xirf(xi)

Find the first four moments ofXifXhas the distribution:

x −21 3

f(x) 1/2 1/4 1/4

NoteM 1 is the mean ofX, andM 2 is used in computing the standard deviation ofX.

Use the formula forMrto obtain:

M 1 =

∑

xif(xi)=− 2

( 1

2

)

+ 1

( 1

4

)

+ 3

( 1

4

)

= 0

M 2 =

∑

x^2 if(xi)= 4

( 1

2

)

+ 1

( 1

4

)

+ 9

( 1

4

)

= 4. 5

M 3 =

∑

x^3 if(xi)=− 8

( 1

2

)

+ 1

( 1

4

)

+ 27

( 1

4

)

= 3

M 4 =

∑

x^4 if(xi)= 16

( 1

2

)

+ 1

( 1

4

)

+ 81

( 1

4

)

= 28. 5

7.41. Prove Theorem 7.10 (Chebyshev’s Inequality): Fork>0,

P(μ−kσ≤X≤μ+kσ)≥ 1 −k^12

By definition

σ^2 =Var(X)=

∑

(xi−μ)^2 pi

Delete all terms from the summation for whichxiis in the interval[μ−kσ, μ+kσ]; that is, delete all terms for which

|xi−μ|≤kσ. Denote the summation of the remaining terms by

∑∗

(xi−μ)^2 pi. Then

[

σ^2 ≥

∑∗

(xi−μ)^2 pi≥

∑∗

k^2 σ^2 pi=k^2 σ^2

∑∗

pi=k^2 σ^2 P(|X−μ|>kσ)

]

=k^2 σ^2 [ 1 −P(|X−μ|≤kσ)]=k^2 σ^2 [ 1 −P(μ−kσ≤X≤μ+kσ)]

Ifσ>0, then dividing byk^2 σ^2 gives

1

k^2 ≥^1 −P(μ−kσ≤X≤μ+kσ) or P(μ−kσ≤X≤μ+kσ)≥^1 −

1

k^2

which proves Chebyshev’s Inequality forσ>0. Ifσ=0, thenxi=μfor allpi>0, and

P(μ−k· 0 ≤X≤μ+k· 0 )=P(X=μ)= 1 > 1 −k^12

which completes the proof.