Answer to Question #235425 in Statistics and Probability for Omniscient

Question

Random sample of size 25 from a normally distributed population with the mean of 4 and variance of 16. Find the point estimate of the population mean if P(X>x)=0.9798?

Answer 1

$$P(X<x)=1-P(Z\leq\dfrac{x-\mu}{\sigma/\sqrt{n}})$$

$$=1-P(Z\leq\dfrac{x-4}{\sqrt{16}/\sqrt{25}})=0.9798$$

$$P(Z\leq1.25(x-4))=0.0202$$

$$1.25(x-4)\approx-2.04963$$

$$x=2.36$$

$x=2.36$

The point estimate of the population mean is 4.