A random sample of size n=36 has a standard deviation of 7. Calculate approximately that the mean of this sample is greater than 44.5. While the mean of the population is 42
"\\sqrt{n}\\frac{\\bar{x}-\\mu}{s}\\sim t_{n-1}=t_{35}\\\\P\\left( \\bar{x}>44.5 \\right) =P\\left( \\sqrt{36}\\frac{\\bar{x}-42}{7}>\\sqrt{36}\\frac{44.5-42}{7} \\right) =\\\\=P\\left( t_{35}>2.14286 \\right) =1-F_{t,35}\\left( 2.14286 \\right) =1-0.9804=0.0196"
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