You draw a random sample of 300 employee records from the personnel file and find that the
average years of service per employee is 6.3, with a standard deviation of 3.0 years.
a. What percentage of the workers would you expect to have more than 9.3 years of service?
b. What percentage would you expect to have more than 5.0 years of service?
Z = (score - mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions from your Z scores.
To calculate the percentage of workers who would have more than a certain number of years of service, we need to use the normal distribution.
a. To find the percentage of workers who would have more than 9.3 years of service, we need to calculate the z-score first. The z-score is calculated using the formula:
z = (x - μ) / σ
Where:
x = 9.3 (number of years of service)
μ = 6.3 (mean)
σ = 3.0 (standard deviation)
Substituting the values, we get:
z = (9.3 - 6.3) / 3.0
z = 1
Next, we need to find the area under the curve to the right of the z-score. We can use a standard normal distribution table, which provides the cumulative probability for different z-scores.
Looking up the z-score of 1 in the table, we find that the cumulative probability is approximately 0.8413. Since we are interested in the area to the right of the z-score, we subtract this probability from 1:
Percentage of workers with more than 9.3 years of service = (1 - 0.8413) * 100 = 15.87%
Therefore, we would expect approximately 15.87% of the workers to have more than 9.3 years of service.
b. To find the percentage of workers who would have more than 5.0 years of service, we follow the same steps as above:
z = (5.0 - 6.3) / 3.0
z = -0.43
Looking up the z-score of -0.43 in the standard normal distribution table, we find that the cumulative probability is approximately 0.3336.
Percentage of workers with more than 5.0 years of service = (1 - 0.3336) * 100 = 66.64%
Therefore, we would expect approximately 66.64% of the workers to have more than 5.0 years of service.