Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score. Multiply by 10,000.
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 (correct to the nearest hundred) would you expect to be earning more than $16.50 per hour?
I don't understand how to solve this.
2 answers
Please someone help me to understand this. Here is what I figured and it is still wrong.
x= 16.50-12.96= 3.54/4.46 (SD)= 0.79
the normal distribution for 0.79 is .285. I then multiplied that by 10,ooo and the answer is: 2,850.
The correct answer is 2200 according to the answer from the test. Where did I go wrong?
x= 16.50-12.96= 3.54/4.46 (SD)= 0.79
the normal distribution for 0.79 is .285. I then multiplied that by 10,ooo and the answer is: 2,850.
The correct answer is 2200 according to the answer from the test. Where did I go wrong?
1 answer