Answer to Question #335349 in Statistics and Probability for Jack

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:

Answer 1

Let $X=$ the mean salary of the sample:$X\sim N(\mu, \sigma^2/n).$

Given $\mu=42500, \sigma=5600, n=25.$

a.

$$P(X<40500)=P(Z<\dfrac{40500-42500}{5600/\sqrt{25}})$$

$$\approx0.0371$$

b.

$$P(X>41000)=1-P(Z\le\dfrac{41000-42500}{5600/\sqrt{25}})$$

$$\approx0.9098$$

c.

$$P(40400<X<45000)$$

$$=P(Z<\dfrac{45000-42500}{5600/\sqrt{25}})$$

$$-P(Z<\dfrac{40400-42500}{5600/\sqrt{25}})$$

$$\approx0.9568$$