Answer to Question #326329 in Statistics and Probability for Kimberly

Question

The mean annual salary of all the frontlines (nurses, medical technologists,

radiologic technologists, phlebotomists) in the Philippines is Php 42,500.

Assume that this is normally distributed with standard deviation Php 5,600.

A random sample of 25 health workers is drawn from this population, find

the probability that the mean salary of the sample is:

Greater than php 41, 000?

Answer 1

We have a normal distribution, $\mu=42500, \sigma=5600, n=25.$

Let's convert it to the standard normal distribution,

$z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$

$z=\cfrac{41000-42500}{5600/\sqrt{25}}=-1.34,$

$P(\bar{X}>41000)=1-P(\bar{X}<41000)=\\ =1-P(Z<-1.34)=$

$=1-0.0901=0.9091$ ​(from z-table).