Question

If $X$ is a random variable such that $E(X)=3$ and $E\left(X^2\right)=13$, use Chebyshev's inequality to determine a lower bound for the probability $P(-2<$ $X<8)$

Short answer

The lower bound for the probability that X lies between -2 and 8 is 0.36.

Step-by-step solution

Find variance of X.

The variance of a random variable $X$ is given by ${\sigma }^2=E\left(X^2\right)-[E(X){]}^2$. Here, we're given that $E(X)=3$ and $E\left(X^2\right)=13$. Hence, ${\sigma }^2=13-3^2=4.$

Standardize the range.

We're looking for the probability that $X$ is between -2 and 8. We can standardize this by calculating how many standard deviations each of these numbers is away from the expected value $E(X)=3$. The standardized range is $(-2-3)/4=-5/4$ to $(8-3)/4=5/4$. Hence, we're looking for the probability that $X$ is less than 5/4 standard deviations away from its mean.

Apply Chebyshev's inequality.

By Chebyshev's inequality, the probability that a random variable $X$ is more than $k$ standard deviations away from its mean is at most $1/k^2$. Hence, the probability that $X$ is within $k$ standard deviations of its mean $E(X)$ is at least $1-1/k^2$. Substituting $k=5/4$ gives the lower bound for the probability as $1-1/(5/4{)}^2=1-16/25=9/25=0.36.$

Interpret the result.

This means that the probability that $X$ lies between -2 and 8 is at least 0.36, according to Chebyshev's inequality.