What percentage of the annual performance reviews in the department take between 25.0 and 30.0 minutes? 1. MBB Problem 7.2 (modified)

What percentage of the annual performance reviews in the department take between 25.0 and 30.0 minutes?
1. MBB Problem 7.2 (modified)

The average (mean) amount of time that a manager at the Cimaron Valley Department of Human Services spends in the annual performance review with an employee is 27.5 minutes, with a standard deviation of 2.5 minutes (normal distribution).

(a) What percentage of the annual performance reviews in the department take between 25.0 and 30.0 minutes?
Solution;
Z= ( x- μ)/ σ, and probability is P(Z= ( x- μ)/σ)
μ=27.5 minutes, σ = 2.5 minutes
The percentage of annual performance reviews between 25.0 and 30.0 minutes can be calculated as ;
P(25.0< > 30.0) = ( (25.0– 27.5) /2.5 )>Z> (30.0− 27.5) /2.5)
=P(-1>Z>1)= P(Z<1)- P(Z<-1)
= 0.84134-0.15866
= 0.68268 or 68.268%
The percentage of the annual performance reviews between 25.0 and 30.0 minutes is 68.268%

(b) Between 22.5 and 32.5 minutes?
Solution

P(22.5< > 32.5) = ( (22.5– 27.5) /2.5 )>Z> (32.5− 27.5) /2.5)
=P(-2>Z>2)= P(Z<2)- P(Z<-2)
= 0.97725 – 0.02275
= 0.9545 or 95.45%
The percentage of the annual performance reviews between 22.5 and 32.50 minutes is 95.45%
(c) Between 20.0 and 35.0 minutes?

P(20.0< > 35.0) = ( (20.0– 27.5) /2.5 )>Z> (35.0 − 27.5) /2.5)
=P(-3>Z>3)= P(Z<3) – P(Z<-3)
= 0.99865 – 0.00135
= 0.9973 or 99.73%
The percentage of the annual performance reviews between 22.5 and 32.50 minutes is 99.73%

(d) You can add two normal distributions together and the new distribution is also a normal distribution with a mean equal to the sum of the two means and a standard deviation equal to the square root of the sum of the two square roots (assuming that the two distributions are independent of each other). So, if the manager did two performance reviews in a row, the two reviews combined would last a mean of 27.5 + 27.5 = 55.0 minutes with a standard deviation of = 3.54 minutes, and if the manager did three reviews in a row, the three reviews combined would last a mean of 27.5 + 27.5 + 27.5 = 82.5 minutes with a standard deviation of = 4.33 minutes. If the manager schedules four performance reviews in a row every day starting at 10:00, what percentage of days would she expect to finish before 12:00.
Solution
For 4 performance reviews in a row
Mean = 27.5 + 27.5+27.5 + 27.5= 110 minutes
minutes
From 10:00 to 12:00, there are 2hours,( 120 minutes)
P(Z< (120− 110) /5) = P(Z<2) = 0.97725 or 97.725%

If the manager schedules four performance reviews in a row every day starting at 10:00, the percentage that would finish before 12:00 is 97.725%

2. MBB Problem 7.4 (modified)
Refer to Problem 7.3 (which we covered in lecture). The head of the Mariposa County Accounting Department wants to establish a standard regarding the length of time that employees can expect to wait to receive reimbursement for professional expenses.
(MBB Problem 7.3According to records kept by MariposaCounty, the average ( mean) amount of time that it takes for employees to be reimbursed for professional expenses incurred in service to the county is 36 days, with a standard deviation of 5 days. The distribution is normal. About 2 months ago, Latisha MacNeilattended a training conference for her job in the County Recreation Department. She filed for reimbursement of expenses that she incurred at the conference 42 days ago, but she has not received payment. What is the probability of receiving reimbursement within 42 days of filing?

Explanation & Answer

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