# Document

```Section 10.5
Let 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 – )2f(x) dx
{x : |x – |  k}
(x – )2f(x) dx 
{x : |x – |  k}
+
{x : |x – | < k}
k2 2 f(x) dx =
{x : |x – |  k}
(x – )2f(x) dx 
k2 2 f(x) dx
{x : |x – |  k}
We now have that 2  k2 2 P(|X – |  k) which implies that
P(|X – |  k)

1
—
k2
This is Chebyshev’s inequality and is stated in Theorem 10.5-1.
We may also write
P(|X – | < k)

1
1 – —2
k
1. Let X be a random selection from one of the first 9 positive integers.
(a) Find the mean and variance of X.
 = E(X) = 5
2
20
= Var(X) = —
3
(b) Find P(|X – 5|  4) .
2
P(|X – 5|  4) = P(X = 1  X = 9) = —
9
(c) 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.
20/3 5
P(|Y – 5|  4) = P[|Y – 5|  4(3/20)1/2(20/3)1/2]  —— = —
16
12
k

2. Let X be a random variable with mean 100 and variance 75.
(a) Find the lower bound on P(|X – 100| < 10) that we get from
Chebyshev’s inequality.
1
1
P(|X – 100| < 10) = P[|X – 100| < (2/3)(53)]  1 – ——–2 = —
(2/3)
4
k

(b) Find what the value of P(|X – 100| < 10) would be, if X had a
U(85 , 115) distribution.
20
2
P(|X – 100| < 10) = P(90 < X < 110) = — = —
30
3
3. Let Y have a b(n, 0.75) distribution.
(a) Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from
Chebyshev’s inequality when n = 12.
When n = 12, E(Y) = 9 , and Var(Y) = 2.25, and
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.)
(b) Find the exact value of P(|Y / n – 0.75| < 0.05) when n = 12.
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) Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from
Chebyshev’s inequality when n = 300.
When n = 300, E(Y) = 225 , and Var(Y) = 56.25, and
P(|Y / n – 0.75| < 0.05) = P(|Y – 225| < 15) =
1
P[|Y – 225| < (2)(7.5)]  1 – —2 = 0.75
2
(d) By using the normal approximation, show that when n = 300, then
P(|Y / n – 0.75| < 0.05) = 0.9464.
```

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