An employer wants to estimate to set a time limit so that 75% of the employees will finish a job on time. Past history has shown that the time required to do the job is normally distributed and has a mean time of 26 minutes with a standard deviation of 5 minutes. How much time should the employer allow employees to finish the job? Round to the nearest minute. Show work.
Use a z-score formula:
z = (x - mean)/sd
Find the z-score that corresponds to 75% using a z-table. Remember that this value will be a positive z-score and above the mean of the distribution. Substitute z, mean, and sd given in the problem, then solve for x.
I hope this will help get you started.
To estimate the time limit for the job, we need to find the value that corresponds to the 75th percentile of the normal distribution.
First, we need to standardize the distribution using the z-score formula:
z = (x - μ) / σ
where:
- x is the value we want to find (time limit)
- μ is the mean time (26 minutes)
- σ is the standard deviation (5 minutes)
Now, we need to find the z-score that corresponds to the 75th percentile. This can be done by finding the z-score using the z-table or using a calculator. With a z-table, we can determine that the z-score for the 75th percentile is approximately 0.674.
Now, we can solve for x in the z-score formula:
0.674 = (x - 26) / 5
Multiply both sides of the equation by 5:
0.674 * 5 = x - 26
3.37 = x - 26
Add 26 to both sides of the equation:
3.37 + 26 = x
x = 29.37
Rounding to the nearest minute, the employer should allow employees to finish the job within 29 minutes.