P(E) will be kσ from its mean?

Hi everyone, In Reading 11 " Sampling and Estimation" 's Practice problems, I encountered a question which I can’t understand:

[question removed by moderator]

Does anybody understand why:

  1. P(2σ)=0.0456 NOT P(2σ)= 0.0228 CUZ according to normal table (z) 1 - P(2σ) = 1 - 0.9772 P(3σ)=0.0026 NOT P(3σ)= 0.0013 CUZ according to normal table (z) 1 - P(3σ) = 1 - 0.9987
  2. P(|X−μ|≥2σ)≤(1/2)2=0.2500 and P(|X−μ|≥3σ)≤(1/3)2=0.1111. What is question B asking ? and what is the specific meaning of the answer ?


A. You’re looking at a 1-tail probability. It can be 2_σ_ or 3_σ_ below the mean, or 2_σ_ or 3_σ_ _ above _ the mean.

B. Chebyshev’s inequality gives a minimum percentage of observations within k_σ_ of the mean for any distribution. Therefore, you can use 1 − Chebyshev’s number to get the maximum percentage of observations that are (or, looking at it another way, the maximum probability of an observation being) more than k_σ_ from the mean.

Thanks you

You’res welcome.