"X_1, X_2, ..., X_n" is a random sample drawn from a normal distribution with "X_i\\thicksim N(\\mu,\\sigma^2)" for all "i". Then
Using this probability distribution, we have
"P(-z_{0.025}<{\\bar{X}-\\mu \\over \\sigma\/\\sqrt{n}}<z_{0.025})=0.95"
95% CI for µ when "\\sigma^2" known and drawing from a normally distributed population:
"\\bar{X}=60, \\sigma=\\sqrt {25}=5, n=16, z_{0.025}=1.96"
"60-1.96{5 \\over \\sqrt{16}}\\le \\mu \\le 60+1.96{5 \\over \\sqrt{16}}"
"57.55\\le \\mu \\le 62.45"
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