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: Identify the mean and variance.
Given
\[ E(X)=2 \] and
\[ Var(X)=4 \] So, the standard deviation is
\[ \sigma=\sqrt{4}=2 \]

Step 2: Rewrite the probability interval around the mean.
The interval is
\[ -4\leq X\leq 8 \] This can be written as
\[ |X-2|\leq 6 \] Since
\[ 6=3\sigma \] we have
\[ P(-4\leq X\leq 8)=P(|X-2|\leq 3\sigma) \]

Step 3: Apply Chebyshev's inequality.
Chebyshev's inequality gives
\[ P(|X-\mu|<k\sigma)\geq 1-\frac{1}{k^2} \] Here,
\[ k=3 \] Therefore,
\[ P(|X-2|\leq 6)\geq 1-\frac{1}{3^2} \] \[ =1-\frac{1}{9} \] \[ =\frac{8}{9} \] \[ =0.888\ldots \] Rounded off to two decimal places,
\[ 0.89 \]

Step 4: Final conclusion.
Hence, the required lower bound is
\[ \boxed{0.89} \]