The owner of kat motel wants to develop a time standard for the task of cleaning a cat cage. in a preliminary study, she observed one of her workers perform this task six times, with the following results: observation 1 2 3 4 5 6 time(seconds) 99 87 90 81 93 90 what is the standard time for this task if the employee worked at a 50 percent faster pace than average, and an allowance of 20 percent of job time is used? 162 seconds 99 seconds 90 seconds

The standard time for this task is 162 seconds ⇒ 1st answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Kat observed one of her workers perform this task six times, with the

 following results

- Observation : 1       2      3      4      5      6

 Time (sec.)    : 99   87    90    81    93    90

- Lets calculate the average of the time

∵ The average time = Total time ÷ number of times

∵ Total = 99 + 87 + 90 + 81 + 93 + 90 = 540

∵ The number of times is 6

∴ The average time = 540 ÷ 6 = 90 seconds

- The standard time = normal time + allowances

∵ Normal time = average time × rating

∵ The average time is 90 seconds

∵ The rate is 50 percent faster pace than average

∴ The rate is (1 + 50/100) = 1.5

∴ The normal time = 90 × 1.5 = 135 seconds

- The allowance of 20 percent of job time is used

∵ The allowance = average time × allowance rate

∵ The allowance is 20%

∴ The allowance = 135 × 20/100 = 135 × 0.2 = 27 seconds

* Lets calculate the standard time

∵ The standard time = normal time + allowances

∵ the normal time is 135 seconds

∵ The allowance time is 27 seconds

∴ The standard time = 135 + 27 = 162 seconds

* The standard time for this task is 162 seconds

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