The time necessary to assemble a discount store display is normally distributed with μ = 5 minutes and σ = 30 seconds. A large number of potential employees are timed assembling a practice display. How much time (in minutes) should the personnel manager allow the potential employees to assemble the display if the store wants only 80% of the people to complete the task?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.80) related to the Z score. Use the Z score in the equation below.
Z = (score-mean)/SD
To find the amount of time the personnel manager should allow the potential employees to assemble the display if the store wants only 80% of the people to complete the task, we need to find the right cutoff value from the normal distribution.
Step 1: Convert the given mean and standard deviation to minutes.
μ = 5 minutes
σ = 30 seconds = 30/60 = 0.5 minutes
Step 2: Determine the z-score for the desired percentage.
Since we want only 80% of the people to complete the task, we need to find the z-score corresponding to the 80th percentile (because the normal distribution is symmetric).
Using a standard normal distribution table or a calculator with z-score functionality, we find that the z-score corresponding to the 80th percentile is approximately 0.8416.
Step 3: Calculate the cutoff value.
The cutoff value can be found using the formula:
cutoff value = μ + (z-score * σ)
cutoff value = 5 + (0.8416 * 0.5)
cutoff value ≈ 5.421 minutes
Therefore, the personnel manager should allow the potential employees an approximate time of 5.421 minutes (or about 5 minutes and 25 seconds) to assemble the display if the store wants only 80% of the people to complete the task.