Question

Problem 1: Suppose is a random variable with mean 64.5 and variance 144. Use Chebyshev’s inequality to approximate the following probabilities.

Answer 1

CHEBYSHEV’S INEQUALITY

Let $X$ be a discrete rv with mean $\mu$ and standard deviation $\sigma$. Then, for any $k\geq1,$

$$P(|X-\mu|\geq k\sigma)\leq{1 \over k^2}$$

Given that $\mu=64.5, \sigma^2=144$

$a)\ \ P(44<X<85)$

$$\sigma=\sqrt{144}=12$$

$${85-44 \over 2}=20.5$$

$$k={20.5 \over 12}$$

$$P(44<X<85)=1-P(|X-\mu|\geq k\sigma)=$$

$$=1-P\bigg(|X-64.5|\geq {20.5 \over 12}\cdot 12\bigg)\geq1-{1 \over (20.5/12)^2}$$

$$P(44<X<85)>0.6573$$

$b) P(36<X<93)$

$$\sigma=\sqrt{144}=12$$

$${93-36 \over 2}=28.5$$

$$k={28.5 \over 12}$$

$$P(36<X<93)=1-P(|X-\mu|\geq k\sigma)=$$

$$=1-P\bigg(|X-64.5|\geq {28.5 \over 12}\cdot 12\bigg)\geq1-{1 \over (28.5/12)^2}$$

$$P(36<X<93)>0.8227$$