(adsbygoogle = window.adsbygoogle || []).push({}); Consider a scenario where samples are randomly selected with replacement. Suppose that the population has a probability distribution with mean µ and variance σ 2 . Each sample Xi , i = 1, 2, . . . , n will then have the same probability distribution with mean µ and variance σ 2 . Now, let us calculate the mean and variance of X_bar: E(X_bar) = 1/n*(E(X1) + E(X2) + · · · + E(Xn)) = 1/n (µ + µ + · · · + µ ) = µ

*X_i is independent random variable.

Hello. I wonder why the expected values of Xi are the same as population averageµ.

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# Sample distribution and expected value.

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