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A methods and measurement analyst wants to develop a time standard for a certain task. In a pre-liminary study, he observed one worker perform the task six times with an average observed time of 20 seconds and a standard deviation of two seconds. What is the standard time for this task if the employee worked at a 20 percent faster pace than average, and an allowance of 25 percent of job time is used? a. 20 seconds.b. 25 seconds.c. 26.7 seconds.d. 30 seconds.e. 32 seconds.

Respuesta :

Answer:

e. 32 seconds.

Explanation:

The computation of the standard time is shown below:

= (Normal time) ÷ (1 - allowance of percent of job time)

where,

Normal time = 20 seconds × 1.2 = 24 seconds

The 1.20 is come from 1 + 20%

And, the allowance of percent of job time is 25%

So, the standard time is

= (24 seconds) ÷ (1 - 0.25)

= (24 seconds) ÷ (0.75)

= 32 seconds

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