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MGT613 - Production / Operations Management Assignment # 2 Solution
MGT613 - Production / Operations Management
Assignment # 2

Find out average population proportion defective which would be the central line on p-chart (p) i.e. total defectives/ sample size*weeks

Sample-- number --No. Of Defectives (b)Sample Proportion defective (p)
1--- 16--- 0.005333333
2--- 13--- 0.004333333
3--- 20--- 0.006666667
4--- 3---- 0.001
5--- 18--- 0.006
6--- 6--- 0.002
7---- 26---- 0.008666667
8--- 9---- 0.003
9--- 8--- 0.002666667
10--- 24--- 0.008
11--- 14---- 0.004666667
12---- 5---- 0.001666667
13---- 12--- 0.004
14---- 19--- 0.006333333
15--- 18--- 0.006
 211--- 0.070333333

a. Find out average population proportion defective which would be the central line on
P-chart (p) i.e. total defectives/ sample size*weeks
p = 211 / (300 * 15)
= 211 / 45000
p = 0.004688889
p = 0.00469 (Approximate)
c. Calculate standard deviation of distribution of proportion defective (σp) using the formula
σp = √ p (1– p)/n. Where n= sample size.
= √ 0.00469 (1 – 0.00469) / 3000
= √ .000001556
= .001247397
σp = .00125
d. Find out upper control limit (UCL) and lower control limit (LCL) using the formula
(UCL p = p + zσ p and LCL p = = p - zσ p )
Use three sigma control limits (z=3)
1) UCL p = p + zσ p
= .00469 + 3 (0.00125)
UCL p = .00844

2) LCL p = p - zσ p
= . 00469 - 3 (0.00125)
LCL p = .00094
e. On the basis of above data construct a p-chart taking sample numbers on x-axis and sample proportion defective (p) on y-axis.




f. Which sample number is showing the highest proportion of defective? Is the process still in control? Analyze the trend in chart.
In the above diagram the sample no.7 is the highest value and this value is out of the upper and the lower limit
UCL p = .00844 and LCL p = .00094
High proportion of the defect is Value = .0087
That’s why we can say that the process is not through the statistical process.