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.

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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