Answer:
Administrative.
Explanation:
The ways in which organizations use the system to provide information for day-to-day decisions about salary, benefits, and recognition programs falls under the <u>Administrative</u> purpose of a performance management system.
Performance management system: It is a system used to evaluate the performance of the employee and rewarding them on the basis of their performance. It also helps in defining the required role of an employee in a given time, which creates transparency in the evaluation process. It serves as a basis for improving employees' knowledge and skills, which motivate employee for better performance.
There are three purposes of performance management:
- Strategic purpose.
- Administrative purpose.
- Developmental purpose.
Strategic purpose: It is defined as how effective performance help the organization to grow.
Administrative purpose: It is defined as how organizations use the system to provide information for day-to-day decisions about salary, benefits, and recognition programs.
Developmental purpose: It is defined as how system helps organization in developing employee´s skill and knowledge.
The inventory system whereby the merchandise inventory account balance is merely a record of the most recent physical inventory count is called the periodic system. The periodic inventory system is a<span> method of accounting for merchandise inventory in which the cost of the inventory sold is determined only at the end of an accounting period.</span>
<span>The updates in this system are made on a </span>periodic<span> basis. </span>
Answer:
Confidence Interval is 139.04 - 142.96
Explanation:
The formula for a confidence interval is as follow:
Mean (Average price) +/- z-score x standard deviation / sqrt(n)
Formula Interpretation:
Mean = $141
z-score for 95% confidence interval = 1.96
standard deviation = $4
n = 16 --> sqrt (n) = 4
By using these inputs, we can calculate the confidence interval as follow:
141 +/- 1.96 x (4/4)
Confidence Interval is 139.04 - 142.96
Answer and Explanation:
Data provided in the question
defect rate i.e.
= 1.50%
the sample size = n = 200
Now

= 0.008595057
Now the 3 sigma control limits is
UCL_p =
+ 35p
= 0.015 + 3 (0.008595057
)
= 0.04078517
LCL_p =
- 35p
= 0.015 - 3 (0.008595057
)
= 0
hence, the 3 sigma control limits are UCL 0.04078517 and LCL 0 respectively