Answer to Question #328650 in Statistics and Probability for moon

Question

the mean serum cholesterol level of a large population of overweight children is 220 mg per deciliter (mg/dl), and the standard deviation is 16.3 mg/gl. If a random sample of 35 over which children is selected, find the probability that the mean will be between 220 and 222 mg/ dl. Assume the serum cholesterol level variable is normally distributed.

Answer 1

$P\left( 220\leqslant \bar{x}\leqslant 222 \right) =P\left( \sqrt{n}\frac{220-\mu}{\sigma}\leqslant \sqrt{n}\frac{\bar{x}-\mu}{\sigma}\leqslant \sqrt{n}\frac{222-\mu}{\sigma} \right) =\\=P\left( \sqrt{35}\frac{220-220}{16.3}\leqslant Z\leqslant \sqrt{35}\frac{222-220}{16.3} \right) =P\left( 0\leqslant Z\leqslant 0.725899 \right) =\\=\varPhi \left( 0.725899 \right) -\varPhi \left( 0 \right) =0.7660-0.5=0.266$