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 (secs) 99 87 90 81 93 90 What is the normal time for this task if the employee worked at a 50 percent faster pace than average

a. 150 seconds

Explanation

Missing word "than average, and an allowance of 10 percent of the workday is used?"

Multiple choice: 150 seconds , 99 seconds , 90 seconds , 168.8 seconds , 100 seconds

σ (Recorded Time) = 99 + 87 + 90 + 81 + 93 + 90 = 540

Observation = n = 6

Performance rating = 100% + 50% = 150%

Observed Time = σ (Recorded Time) / n

Observed Time = 540 * 6

Observed Time = 90

Normal time = Observed Time * Performance rating

Normal time = 90 * (150 / 100)

Normal time = 90 * 1.5

Normal time = 135

Standard Time = Normal time / (1 - Allowance factor)

Standard Time = 135 / (1 - (10 / 100))

Standard Time = 135 / (1 - 0.1)

Standard Time = 135 / 0.9

Standard Time = 150 seconds.

answer; /// i believe the correct answer is first in, first out(fifo); //