 # Project Management Course

User Generated

pjunegba78

## Description

PROJECT QUALITY MANAGEMENT

Text: Managing for Quality and Performance Excellence, Ninth Edition, Chapters 6-13

Author: James r. Evans & William M. Lindsay.

Each question should be answered in 200 or more words when it's not an equation. Please see attached files.

### Unformatted Attachment Preview

Purchase answer to see full attachment User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service. Project quality management
Name
Institution
Professor
Course
Date

1

2

PROJECT QUALITY MANAGEMENT
Part I

1. The Turkumann Rug Company buys medium grade carpet in 100-ft rolls. The average
number of defects per turn is 1.8. Assuming that these data follow a Poisson distribution:
a. What is the probability of finding precisely seven defects in a carpet roll chosen at
random?

Therefore the probability of getting exactly 7 =
𝑃(𝑋 = 𝑥) =

𝑃(𝑋 = 𝑥) =

𝑒 −1.8 1.87
7!

2.7183−1.8 1.87
7!

Therefore the probability of finding exactly seven defectives is
0.16529∗61.222
5040

=

0.0020078

b. What is the probability of getting 4 or fewer defectives?
P(0)+P(1)+P(2)+P(3)+P(4)
2.7183−1.8 1.80 2.7183−1.8 1.81 2.7183−1.8 1.82 2.7183−1.8 1.83 2.7183−1.8 1.84
+
+
+
+
=
0!
1!
2!
3!
4!
= 0.1652+0.2975+0.2677+0.1607+0.0723
= 1.0957-1=0.0957
0.0957

2. The dimension of a machined part has a nominal specification of 11.9 cm. The process
that produces that part can be controlled to have a mean value equal to this specification,
but has a standard deviation of 0.05 cm. What is the probability that a part will have a
dimension?
a. Exceeding 12 cm?

3

PROJECT QUALITY MANAGEMENT

1

1 12−11.9
)
0.05

F(x) = 0.05∗√2∗3.142 𝑒 −2(

= 0.2222*0.3678 = 0.0817
b. Between 11.9 and 11.95 cm?

1

1 11.95−11.9
)
0.05

F(x) = 0.05∗√2∗3.142 𝑒 −2(

1

1 11.9−11.9
)
0.05

- F(x) = 0.05∗√2∗3.142 𝑒 −2(

0.2222 – 0.2222*0.6065 = 0.0874
c. Less than 11.83
1

1 11.83−11.9
)
0.05

F(x) = 0.05∗√2∗3.142 𝑒 −2(

= 0.2222*2.0137 = 0.4474

3. The hypothesis testing
Ho : 0.87
H1 ; 0.13

4. A blueprint specification for the thickness of a refrigerator part at Refrigeria, Inc. is
0.300 ±0.025 cm. It costs \$25 to scrap a part that is outside the specifications.
a. Determine the Taguchi loss function for this situation.
L(X) = 𝑘(𝑥 − 𝑁)2
b. If the process deviation from a target can reduce to 0.015 cm, what is the Taguchi Loss?
L(X) = 𝑘(𝑥 − 𝑁)2
Where k= c/𝑑2
C is the loss associated with sp limit
D deviation of specification from target value

4

PROJECT QUALITY MANAGEMENT

25= k(0.025)2 this is because the tolerance given as the deviation value thus this is the Taguchi
function
1failure

1

5. Failure rate=number of hours= 18000 = 0.000055

6. The overall reliability of the system is given by
a.

Rs = P1*P2*P3 = 0.998*0.992*0.978 = 0.968

b. The overall reliability is 0.968 meaning that the process goes on well and customer will serve
well

𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑛𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒

7. NPU =

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑒𝑑
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒

38

= 100 = 0.38
62

TY= 𝑡𝑜𝑡𝑎𝑙 𝑢𝑛𝑖𝑡𝑠 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 100 = 0.62
RTY = TY1 * TY2 * TY3
= 0.83*0.90*0.92 = 0.68724
Proportion nonconforming =

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑛𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑖𝑛𝑔 𝑓𝑜𝑢𝑛𝑑
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑒𝑑

31

= 100 = 0.31

8. Suppose that a refrigeration process at Cool Foods, Ltd. has a normally distributed
output
with a mean of 25.0 and a variance of 1.44.
a. If the specifications are 25.0 ± 3.25, compute Cp, Cpk, and Cpm· Is the process capable
and centered?
𝑍=

𝑥−𝑢

3.25

𝑠𝑑

1.2

=

= 2.708

Cpk = 2.78
As Cpk is greater than 2, then it is capable
Cp = ( USL-LSL)

PROJECT QUALITY MANAGEMENT
If a process is perfectly centered, then we know that the (USL... ### Review Anonymous
Really great stuff, couldn't ask for more. Studypool 4.7 Trustpilot 4.5 Sitejabber 4.4