Given:
"\\mu=10\\\\\\sigma=2\\\\n=100\\\\"
ind the probability that the mean of a sample of size 100 will differ from the population mean 10 by at most 0.3 unit, that is, is either more than 9.7 or less than 10.3The corresponding z-value needed to be computed
Formula: "Z= \\frac{\\overline{X}-\\mu}{\\frac{\\sigma }{\\sqrt{n}}}" = "\\frac{9.7-10}{\\frac{2 }{\\sqrt{100}}}" = -1.5
Z= "\\frac{10.3-10}{\\frac{2 }{\\sqrt{100}}}" =1.5
Using z table,"Pr(9.7\u2264 \n\\overline{X}\n \u226410.3)" "=Pr(\u22121.5\u2264Z\u22641.5)"
"=Pr(Z\u22641.5)\u2212Pr(Z\u2264\u22121.5)"
=0.9332−0.0668=0.8664
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