Based on a careful work study in the Richard Dulski$$ Corp., the results shown in the following table have been$$ observed:

Element

Observations$($minutes per$$ cycle$)$

Performance

Rating

$1$

$2$

$3$

$4$

$5$

Prepare daily reports

$36$

$39$

$33$

$41$

$40$

$120\%$

Photocopy results

$13$

$9$

$37$

$15$

$14$

$110\%$

Label and package reports

$2$

$4$

$5$

$4$

$5$

$85\%$

Distribute reports

$14$

$18$

$22$

$17$

$45$

$85\%$

Photocopying the results during observation $3$ was influenced by the photocopying machine being broken. Distributing the reports during observation $5$ was influenced by a power outage. Both of these are included as delays in the allowance factor.

The allowance factor for this type of work is given to be $18\%.$

Part $2$

To determine the$$ average, one needs first to drop the outliers for which an exception exists. For the given$$ data, the average time for the$$ "prepare daily$$ reports" operation$=$

$$

$37\mathrm{.}80$ minutes. $($Round your response to two decimal$$ places.$)$

Part $3$

Normal Time$=$ Average TimetimesPerformance Rating Factor

The normal time for the$$ "prepare daily$$ reports" operation$=$

$$

$45\mathrm{.}36$ minutes. $($Round your response to two decimal$$ places.$)$

Part $4$

The average time for the$$ "photocopy results" operation$=$

$$

$12\mathrm{.}75$ minutes. $($Round your response to two decimal$$ places.$)$

Part $5$

Normal Time$=$ Average TimetimesPerformance Rating Factor

The normal time for the$$ "photocopy results" operation$=$

$$

$14\mathrm{.}03$ minutes. $($Round your response to two decimal$$ places.$)$

Part $6$

The average time for the$$ "label and package$$ reports" operation$=$

$$

$4\mathrm{.}00$ minutes. $($Round your response to two decimal$$ places.$)$

Part $7$

Normal Time$=$ Average TimetimesPerformance Rating Factor

The normal time for the$$ "label and package$$ reports" operation$=$

$$

enter your response here minutes. $($Round your response to two decimal$$ places.$)$