MGT613 - Production / Operations Management Assignment # 2 Solution

MGT613 - Production / Operations Management

Assignment # 2

b. Find out sample proportion defective (p) i.e. no. of defectives per week / Sample size

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 (σ_p) using the formula

σ_p = √ p (1– p)/n. Where n= sample size.

= √ 0.00469 (1 – 0.00469) / 3000

= √ .000001556

= .001247397

σ_p = .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.

Note...plzz dont copy past material:

Regards

Muhammad azeem

vuguide.tk