Question

What is the computed z of the problem stated below? 

The average monthly salary for a call center representative in the Philippines is P 21,700 with a standard deviation of P 6,000. Find the probability that a group of 64 randomly selected call center representatives has an average salary higher than P 22,000 per month.

Answer 1

Solution:

Let'a denote given values;

$\mu=21700$ P - population mean;

$\chi=22000$ P - sample mean;

$\sigma=6000$ P - standard deviation;

$n=64$ - sample number.

So, find z score:

$z=\frac{\Chi-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{22000-21700}{\frac{6000}{\sqrt{64}}}=0.4;$

$z$ score of 0.4 equal $0.6554;$

We need to find probability of higher than 22000. So $1-0.6554=0.3446$

$P=0.3446=34.46\%$

Answer:

$z=0.4;$

$P=34.46\%.$