Question

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

Answer 1

$\sqrt{n}\frac{\bar{x}-\mu}{s}\sim t_{n-1}=t_{35}\\P\left( \bar{x}>44.5 \right) =P\left( \sqrt{36}\frac{\bar{x}-42}{7}>\sqrt{36}\frac{44.5-42}{7} \right) =\\=P\left( t_{35}>2.14286 \right) =1-F_{t,35}\left( 2.14286 \right) =1-0.9804=0.0196$