"\\mu=20.8,\\sigma=1.9,\n\nn=84"
Using central limit theorem-
"\\bar{X}" ~"~N(\\mu,\\dfrac{\\sigma}{\\sqrt{n}})"
So mean of sampling distribution "\\bar{x}=\\mu=20.8"
Standard deviation of samples "s=\\dfrac{\\sigma}{\\sqrt{n}}=\\dfrac{1.9}{\\sqrt{84}}=0.2073"
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