A methods and measurement analyst wants to develop a time standard for a certain task. In a preliminary study, he observed one worker perform the task six times with an average observed time of 20 seconds and a standard deviation of 2 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 the workday is used

32 seconds

Explanation:

According to the scenario, computation of the given data are as follows,

Average observed time = 20 seconds

so, if the employee worked at 20% faster pace, then

Performance rating = 1.20

So, Normal time = 20 seconds × 1.20 = 24 seconds

Here, Allowance = 25%

Now, we can calculate the standard time by using following formula,

Standard time = Normal time ÷ ( 1 - Allowance)

By putting the value, we get

Standard time = 24 seconds ÷ ( 1 - 0.25)

= 24 ÷ 0.75

= 32 seconds

wheres the chart? so i can answer.