n = 16
Mean μ = 76
Standard deviation σ = 12.5
"\u03bc\\bar{X} = \u03bc = 76 \\\\\n\n\u03c3\\bar{X} = \\frac{\u03c3}{\\sqrt{n}} = \\frac{12.5}{\\sqrt{16}} = 3.125 \\\\\n\nP(\\bar{X} < 83) = P(\\frac{\\bar{X} \u2013 \u03bc\\bar{X}}{\u03c3\\bar{X}} < \\frac{83 \u2013 76}{3.125}) \\\\\n\nP(z < 2.24) = 0.9874"
Answer: 98.74 %
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