Question

Estimating production time.A widely used technique for estimating the length of time it takes workers to produce a product is the time study. In a time study, the task to be studied is divided into measurable parts, and each is timed with a stopwatch or filmed for later analysis. For each worker, this process is repeated many times for each subtask. Then the average and standard deviation of the time required to complete each subtask are computed for each worker. A worker’s overall time to complete the task under study is then determined by adding his or her subtask-time averages (Gaither and Frazier, Operations Management, 2001). The data (in minutes) given in the table are the result of a time study of a production operation involving two subtasks.

	Worker A		Worker B
Repetition	Subtask 1	Subtask 2	Subtask 1	Subtask 2
1	30	2	31	7
2	28	4	30	2
3	31	3	32	6
4	38	3	30	5
5	25	2	29	4
6	29	4	30	1
7	30	3	31	4

a.Find the overall time it took each worker to complete the manufacturing operation under study.

b.For each worker, find the standard deviation of the seven times for subtask 1.

c.In the context of this problem, what are the standard deviations you computed in part bmeasuring?

d.Repeat part b for subtask 2.

e.If you could choose workers similar to A or workers similar to B to perform subtasks 1 and 2, which type would you assign to each subtask? Explain your decisions on the basis of your answers to parts a–d.

Short answer

The overall time taken by B is more than A.

Step-by-step solution

Finding the overall time taken to complete the work

Worker A

$\begin{array}{rcl}\mathbf{Meanforsubtask}\mathbf{1}& \mathbf{=}& \frac{\mathbf{30}\mathbf{+}\mathbf{28}\mathbf{+}\mathbf{31}\mathbf{+}\mathbf{38}\mathbf{+}\mathbf{25}\mathbf{+}\mathbf{29}\mathbf{+}\mathbf{30}}{\mathbf{7}}\\ & \mathbf{=}& \frac{\mathbf{211}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{30}\mathbf{.}\mathbf{14}\\ & & \\ \mathbf{Meanforsubtask}\mathbf{2}& \mathbf{=}& \frac{\mathbf{2}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{3}}{\mathbf{7}}\\ & \mathbf{=}& \frac{\mathbf{21}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{3}\\ & & \\ \mathbf{Overalltimetaken}& \mathbf{=}& \mathbf{Meanforsubtask}\mathbf{1}\mathbf{+}\mathbf{Meanforsubtask}\mathbf{2}\\ & \mathbf{=}& \mathbf{30}\mathbf{.}\mathbf{14}\mathbf{+}\mathbf{3}\\ & \mathbf{=}& \mathbf{33}\mathbf{.}\mathbf{14}\\ & & \end{array}$

Therefore, the overall time taken by Worker A to complete the work under study is 33.14 mins.

Worker B

$\begin{array}{rcl}\mathbf{Meanforsubtask}\mathbf{1}& \mathbf{=}& \frac{\mathbf{31}\mathbf{+}\mathbf{30}\mathbf{+}\mathbf{32}\mathbf{+}\mathbf{30}\mathbf{+}\mathbf{29}\mathbf{+}\mathbf{30}\mathbf{+}\mathbf{31}}{\mathbf{7}}\\ & \mathbf{=}& \frac{\mathbf{213}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{30}\mathbf{.}\mathbf{43}\\ & & \\ \mathbf{Meanforsubtask}\mathbf{2}& \mathbf{=}& \frac{\mathbf{7}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{6}\mathbf{+}\mathbf{5}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{4}}{\mathbf{7}}\\ & \mathbf{=}& \frac{\mathbf{29}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{4}\mathbf{.}\mathbf{14}\\ & & \\ \mathbf{Overalltimetaken}& \mathbf{=}& \mathbf{Meanforsubtask}\mathbf{1}\mathbf{+}\mathbf{Meanforsubtask}\mathbf{2}\\ & \mathbf{=}& \mathbf{30}\mathbf{.}\mathbf{43}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{14}\\ & \mathbf{=}& \mathbf{34}\mathbf{.}\mathbf{57}\\ & & \end{array}$

Therefore, the overall time taken by Worker A to complete the work under study is 34.57 mins.

Calculating the standard deviation for each worker for subtask 1

Worker A			Worker B
x	$(\mathrm{x}-\overline{\mathrm{x}})$	$(\mathrm{x}-\overline{\mathrm{x}}{)}^2$	y	$(\mathrm{y}-\overline{\mathrm{y}})$	$(\mathrm{y}-\overline{\mathrm{y}}{)}^2$
30	-0.14	0.0196	31	0.57	0.3249
28	-2.14	4.5796	30	-0.43	0.1849
31	0.86	0.7396	32	1.57	2.4649
38	7.86	61.7796	30	-0.43	0.1849
25	-5.14	26.4196	29	-1.43	2.0449
29	-1.14	1.2996	30	-0.43	0.1849
30	-0.14	0.0196	31	0.57	0.3249
SUM	0	94.8572	SUM	0	5.7143

localid="1668432137286" $\begin{array}{rcl}\mathbf{VarianceforworkerA}& \mathbf{=}& \frac{\mathbf{\sum }\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}{\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}}\\ & \mathbf{=}& \frac{\mathbf{94}\mathbf{.}\mathbf{8572}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{13}\mathbf{.}\mathbf{55}\\ & & \\ \mathbf{Standarddeviation}& \mathbf{=}& \sqrt{\mathbf{Variance}}\\ & \mathbf{=}& \sqrt{\mathbf{13}\mathbf{.}\mathbf{55}}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{68}\\ & & \\ \mathbf{VarianceforworkerB}& \mathbf{=}& \frac{\mathbf{\sum }\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}{\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{.}\mathbf{7143}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{816}\\ & & \\ \mathbf{Standarddeviation}& \mathbf{=}& \sqrt{\mathbf{Variance}}\\ & \mathbf{=}& \sqrt{\mathbf{0}\mathbf{.}\mathbf{816}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{9}\\ & & \end{array}$

Standard deviations computed in part b measuring

Worker A's standard deviation for completing subtask 1 is 3.68 minutes.

The standard deviation for completing subtask 1 for Worker B is 0.9 minutes.

Computing the standard deviation for each worker for subtask 2

Worker A			Worker B
x	$(\mathrm{x}-\overline{\mathrm{x}})$	$(\mathrm{x}-\overline{\mathrm{x}}{)}^2$	y	$(\mathrm{y}-\overline{\mathrm{y}})$	$(\mathrm{y}-\overline{\mathrm{y}}{)}^2$
2	-1	1	7	2.86	8.1796
4	1	1	2	-2.14	4.5796
3	0	0	6	1.86	3.4596
3	0	0	5	0.86	0.7396
2	-1	1	4	-0.14	0.0196
4	1	1	1	-3.14	9.8596
3	0	0	4	-0.14	0.0196
SUM	0	4	SUM	0	26.8527

localid="1668431443790" $\begin{array}{rcl}\mathbf{VarianceforworkerA}& \mathbf{=}& \frac{\mathbf{\sum }\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}{\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}}\\ & \mathbf{=}& \frac{\mathbf{4}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{57}\\ & & \\ \mathbf{Standarddeviation}& \mathbf{=}& \sqrt{\mathbf{Variance}}\\ & \mathbf{=}& \sqrt{\mathbf{0}\mathbf{.}\mathbf{57}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{75}\\ & & \\ \mathbf{VarianceforworkerB}& \mathbf{=}& \frac{\mathbf{\sum }\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}{\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}}\\ & \mathbf{=}& \frac{\mathbf{26}\mathbf{.}\mathbf{8527}}{\mathbf{7}}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{8361}\\ & & \\ \mathbf{Standarddeviation}& \mathbf{=}& \sqrt{\mathbf{Variance}}\\ & \mathbf{=}& \sqrt{\mathbf{3}\mathbf{.}\mathbf{8361}}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{96}\\ & & \end{array}$

Worker A's standard deviation for completing subtask 2 is 0.75 minutes.

The standard deviation for completing subtask 2 for Worker B is 1.96 minutes.

Determining the distribution of work

The overall time taken by B is more than A.

The standard deviation for subtask 1 is more significant for Worker A than B. Greater standard deviation implies that A's time to complete subtask 1 varies more, and hence there is no certainty as to when the work will get done. Therefore, type B workers should be given subtask 1.

Type A workers should be given subtask 2 because their standard deviation is smaller than worker Bs, implying that there will be a certainty that the work will get done quickly and the time taken will not vary much.