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.
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.
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?
To solve this problem, we can use the concept of the normal distribution and the z-score.
The average hourly wage of employees is given as $12.96, and the standard deviation is $4.46. We want to find the number of employees earning more than $16.50 per hour.
First, we need to find the z-score, which indicates how many standard deviations away from the mean the value of $16.50 is. The formula for calculating the z-score is:
z = (x - μ) / σ
where x is the value we are interested in (in this case, $16.50), μ is the mean, and σ is the standard deviation.
Plugging in the values, we get:
z = (16.50 - 12.96) / 4.46
z ≈ 0.796
Next, we need to find the proportion of values above this z-score. We can do this by looking up the corresponding value in the z-table or by using a statistical calculator.
Looking up the z-score of 0.796 in the z-table, we find that the proportion is 0.7867 or 78.67%.
Finally, to find the number of employees earning more than $16.50 per hour in a randomly selected sample of 10,000 employees, we multiply the proportion by the sample size:
number of employees = proportion * sample size = 0.7867 * 10,000
Rounding to the nearest hundred:
number of employees ≈ 7,870
Therefore, we would expect approximately 7,870 employees to be earning more than $16.50 per hour in a sample of 10,000 such employees.