Exercise · ZW-LAW-OF-LARGE-NUMBERS-010 · ★★★ Advanced
With \(\sigma^2=1\), which \(n\) suffices, by Chebyshev, so that \(P(|\bar X_n-\mu|\ge0.1)\le0.05\)?
Choose one option.
Show answer
- C. \(n\ge2000\)
Explanation
We need \(\dfrac{1/n}{0.01}\le0.05\), that is, \(n\ge\dfrac{1}{0.05\cdot0.01}=2000\).
Review the concept: Law of large numbers
The law of large numbers states that the average \(\bar X_n\) of \(n\) independent observations of a variable \(X\) approaches its expectation \(\mu\) as \(n\) grows.
\[\bar X_n=\dfrac1n\sum_{i=1}^{n}X_i\ \xrightarrow[n\to\infty]{}\ \mu\]
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