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CSS Statistics QUESTION #1994
Question 1
If sampling is done with replacement from a population with mean \(\mu = 4\) and variance \(\sigma^2 = \frac{20}{9}\), then by Chebyshev's theorem, the probability that the sample mean \(\bar{X}\) (for \(n = 36\)) lies between 3.26 and 4.74 is AT LEAST:
  • \(0.75\)✔️
  • \(0.95\)
  • \(0.50\)
  • \(0.68\)
Correct Answer Explanation
Here \(\sigma_{\bar{X}}^2 = \sigma^2/n = (20/9)/36 = 20/324\), so \(\sigma_{\bar{X}} = \sqrt{20}/18\). The interval \((3.26, 4.74)\) is \(\mu \pm 0.74\). Then \(k = 0.74/\sigma_{\bar{X}}\). Chebyshev gives \(P(|\bar{X}-\mu|\leq k\sigma_{\bar{X}}) \geq 1 - 1/k^2\). With \(k^2 = 4\), the lower bound is \(1 - 1/4 = 0.75\).