Question

Table 1: Customer Wait Time (in minutes) at the Downtown Lube & Oil
	SAMPLE
	Day 1	Day 2	Day 3	Day 4
Wait Time (minutes)	25	28	28	28
	28	33	30	37
	21	24	26	39
	32	27	28	38
	28	37	34	36
	22	29	36	43
	34	29	28	33
	25	30	34	30
	24	27	25	36
	29	33	44	45

Sometimes both x-bar chart and p-chart are available to monitor the same process. In Table 1 above, you collect measurement data (wait times in minutes) to create an x-bar chart. You can also collect count data (pass/fail data) instead to create a p-chart. As you are curious about how it works, you decide to create a p-chart, too.

First, you convert the measurement data in Table 1 to count data using the following criteria:

	Criteria
Not defective (i.e., wait time is appropriate)	25 minutes ≤ wait time ≤ 35 minutes
Defective (i.e., wait time is too short or too long)	wait time < 25 minutes or wait time > 35 minutes 1)

For example, the third wait time on Day 1 in Table 1 is 21 minutes. This is less than 25 minutes. Therefore, you count this as a defective. Table 2 shows the number of defectives.

Table 2: Number of Defectives

	SAMPLE
	Day 1	Day 2	Day 3	Day 4
Number of Defectives	3	2	2	7

Question 1

Based on the Table 2, determine the probability of defectives (p).

1. p=0.35
2. p=1.40
3. p=3.50
4. p=5.60
5. p=14.00

Question 2

Continued from Question 1. Determine the standard deviation of p (σ_p). Round your answer to two decimal places.

1. σ_p = 0.02
2. σ_p = 0.08
3. σ_p = 0.15
4. σ_p = 0.24
5. σ_p = 0.37

Question 3

Continued from Question 14. Determine three-sigma (i.e., z=3) upper and lower control limits for a p-chart. Round your answers to two decimal places.

1. LCL=0.28, UCL=2.52
2. LCL=0, UCL=1.07
3. LCL=0, UCL=0.80
4. LCL=0.12, UCL=0.58
5. LCL=0.28, UCL=0.42

Question 4

Based on the p-chart, is the process in control? If not, which day(s) is not in control?

1. Yes. The process is in control.
2. No. Day 4 is not in control.
3. No. Day 3 and Day 4 are not in control.
4. No. Day 2, Day 3 and Day 4 are not in control.
5. No. All days are not in control.

* I assume that customer wait times (population) are normally distributed with a mean of 30 minutes and a standard deviation of 5 minutes. This implies that the mean of customer wait times (sample mean) is normally distributed with a mean of 30 minutes and a standard deviation of 510 minutes, where 10 is the number of observations in each sample (i.e., sample size). Then, the upper and lower thresholds are calculated as 30±3510. *

Answer 1

The following table shows all the calculations -

	Day 1	Day 2	Day 3	Day 4	Total
Number of defectives	3	2	2	7
Fraction defective(pi)	0.3	0.2	0.2	0.7	1.40
(pi - )	-0.05	-0.15	-0.15	0.35	0.0
(pi - )2	0.0025	0.0225	0.0225	0.1225	0.17

The mean of fraction defective , = (0.3 + 0.2 + 0.2 + 0.7) / 4 = 0.35

Answer 1:
The probability of defectives , p = 0.35 (OPTION A)

Answer 2 :

Hence , OPTION D is correct

Answer 3 :

Hence , OPTION B is correct

Answer 4 :
Since , all the values lie within the control limits , the process is in - control

Hence , OPTION A is correct.