Section 10.5Let X be any random variable with (finite) mean and (finite) variance 2. We shall assume X is a continuous type random variable with p.d.f. f(x), but what follows applies to a discrete type random variable where integral signs are replaced by summation signs and the p.d.f. is replaced by a p.m.f. For any k 1, we observe that

2 = E[(X – )2] =

–

(x – )2f(x) dx =

{x : |x – | k}

(x – )2f(x) dx +

{x : |x – | < k}

(x – )2f(x) dx

{x : |x – | k}

(x – )2f(x) dx

{x : |x – | k}

k2 2 f(x) dx =

{x : |x – | k}

k2 2 f(x) dx

We now have that 2 k2 2 P(|X – | k) which implies that

1P(|X – | k) —

k2

This is Chebyshev’s inequality and is stated in Theorem 10.5-1.

We may also write

1P(|X – | < k) 1 – —

k2

1.

(a)

(b)

(c)

Let X be a random selection from one of the first 9 positive integers.

Find the mean and variance of X.

= E(X) = 2 = Var(X) =520— 3

Find P(|X – 5| 4) .

If Y is any random variable with the same mean and variance as X, find the upper bound on P(|Y – 5| 4) that we get from Chebyshev’s inequality.

P(|X – 5| 4) = P(X = 1 X = 9) = 2— 9

P(|Y – 5| 4) =P[|Y – 5| 4(3/20)1/2(20/3)1/2] 20/3 5—— = — 16 12

k

2.

(a)

(b)

Let X be a random variable with mean 100 and variance 75.

Find the lower bound on P(|X – 100| < 10) that we get from Chebyshev’s inequality.

P(|X – 100| < 10) = P[|X – 100| < (2/3)(53)] 1

1 – ——– = (2/3)2

Find what the value of P(|X – 100| < 10) would be, if X had a U(85 , 115) distribution.

1— 4

P(|X – 100| < 10) =20— =30P(90 < X < 110) =

2— 3

k

3.

(a)

(b)

Let Y have a b(n, 0.75) distribution.

Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from Chebyshev’s inequality when n = 12.

Find the exact value of P(|Y / n – 0.75| < 0.05) when n = 12.

When n = 12, E(Y) = , and Var(Y) = , and9 2.25

P(|Y / n – 0.75| < 0.05) = P(|Y – 9| < 0.6) =

P[|Y – 9| < (0.4)(1.5)] (A lower bound cannot be found, since 0.4 < 1.)

When n = 12, P(|Y / n – 0.75| < 0.05) = P(|Y – 9| < 0.6) =

P(Y = 9) = 0.6488 – 0.3907 = 0.2581

(c)

(d)

Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from Chebyshev’s inequality when n = 300.

By using the normal approximation, show that when n = 300, then P(|Y / n – 0.75| < 0.05) = 0.9464.

When n = 300, E(Y) = , and Var(Y) = , and225 56.25

P(|Y / n – 0.75| < 0.05) = P(|Y – 225| < 15) =

P[|Y – 225| < (2)(7.5)] 11 – — = 0.75 22