Question

Let \(X\) be a continuous random variable with mean \(2\) and variance \(4\). Using Chebyshev's inequality, the lower bound for \(P(-4\leq X\leq 8)\) equals (rounded off to two decimal places).

Hint:

Chebyshev's inequality gives \(P(|X-\mu|\leq k\sigma)\geq 1-\frac{1}{k^2}\), regardless of the distribution of \(X\).

Answer 1

Correct Answer: 0.89

Step 1: Note the spread.
Given mean $2$ and variance $4$, the standard deviation is $\sigma=2$.

Step 2: Centre the interval.
The range $-4\le X\le8$ is the same as $|X-2|\le6$, and $6=3\sigma$.

Step 3: Apply Chebyshev.
With $k=3$, $P(|X-2|\le3\sigma)\ge1-\frac1{k^2}=1-\frac19=\frac89$.

Step 4: Evaluate.
$\frac89=0.888\ldots$

Step 5: Round.
\[ \boxed{0.89} \]