Question

The average hourly wage of employees in a large nationwide industry is $12.96. Assume the variable is normally distributed, with a standard deviation of $4.46. In a randomly selected sample of 10,000 such employees, how many of them would you expect to be earning less than $6.18 per hour?

Answer 1

find the z-score

z = (6.18 - 12.96)/4.46

look up z-score in probability table to find probability of someone earning less than 6.18
multiply this by 10,000