Section AQuestion 1It is Compulsory.Section BQuestion 2Paragraph 1: what is quality management?➢ outline how juran’s is influenced modern quality management.➢ their impact on quality management.Obs: t

Question 1:

Alan’s consultancy generated preliminary data from a client’s process, as presented in Table 1 below.

(a) Using Excel, construct a frequency distribution and histogram for the data. From what type of distribution might you suspect the data are drawn? (10 Marks)

The data is drawn from a normal distribution.  The histogram forms a bell shape meaning that most of the observed data is clustered near the mean, while the data become less frequent when farther away from the mean

Bin

Frequency

1.7 – 2.2

5

2.2 – 2.7

12

2.7 – 3.2

16

3.2 – 3.7

22

3.7 – 4.2

23

4.2 – 4.7

13

4.7 – 5.2

8

5.2 – 5.7

1

(b) From the data below, calculate summary statistics using the Excel Descriptive Statistics tool. Interpret the results of the mean, standard deviation, kurtosis and skewness for managerial purposes. (15 Marks)

The mean for managerial purposes was 3.59 with a standard deviation of 0.81. The low standard deviation is an indication that the data is clustered around the mean. Since the skewness is between -0.5 and 0.5, the data can be described to be fairly symmetrical. Additionally, the kurtosis value (0.29), shows that the data is normally distributed.

Descriptive statistics

Mean

3.591

Standard Error

0.08088

Median

3.65

Mode

3.6

Standard Deviation

0.808802

Sample Variance

0.654161

Kurtosis

-0.29187

Skewness

-0.24044

Range

3.6

Minimum

1.7

Maximum

5.3

Sum

359.1

Count

100

(c) Suppose that a process at Ali Automation has a normally distributed output with a mean of 50.0 cm and a variance of 3.61 cm.

(i) If the specifications are 51.0 ± 3.75 cm., compute Cp, Cpu, Cpl and Cpk. (10 Marks)

Cp =    =   = 0.66

Cpu =      =       = 0.83

Cpl =       =      = 0.48

CpK = min (Cpl, Cpu)  = min (0.48, 0.83)  =0.48

(ii) Suppose the mean shifts to 52.0, but the variance remains unchanged. Recompute and interpret these process capability indexes. (15 Marks)

Cp =    =   = 0.66

Cpu =      =       = 0.48

Cpl =       =      = 0.83

CpK = min (Cpl, Cpu)  = min (0.48, 0.83)  =0.48

Cpk is 0.48. Since it is less than 1, it is considered poor and the process is not capable of repeatedly producing parts that meet specifications. Therefore, the process may need improvement.

d) Belgrave Swim Club is trying to calibrate its chlorine pump to ensure that the right amount of chlorine (1.0–1.5 ppm of free chlorine) is mixed into the water. Thirty samples of four readings at random times during the week were taken, as presented in Table 2 below.

 (i) Compute the mean and range of each sample, calculate control limits, and plot them on x� and R control charts. (45 Marks)

x bar chart

r chart

Obs 1

Obs 2

Obs 3

Obs 4

X-bar mean

r range

LCL

CL

UCL

LCL

CL

UCL

1.22

1.32

1.21

1.35

1.275

0.14

1.171509

1.241

1.310991

0

0.096

0.218311

1.36

1.32

1.31

1.37

1.340

0.06

1.171509

1.241

1.310991

0

0.096

0.218311

1.22

1.25

1.33

1.28

1.270

0.11

1.171509

1.241

1.310991

0

0.096

0.218311

1.26

1.25

1.22

1.21

1.235

0.05

1.171509

1.241

1.310991

0

0.096

0.218311

1.29

1.23

1.34

1.15

1.253

0.19

1.171509

1.241

1.310991

0

0.096

0.218311

1.31

1.25

1.24

1.26

1.265

0.07

1.171509

1.241

1.310991

0

0.096

0.218311

1.24

1.26

1.24

1.18

1.230

0.08

1.171509

1.241

1.310991

0

0.096

0.218311

1.28

1.26

1.19

1.29

1.255

0.10

1.171509

1.241

1.310991

0

0.096

0.218311

1.28

1.22

1.27

1.2

1.243

0.08

1.171509

1.241

1.310991

0

0.096

0.218311

1.14

1.19

1.18

1.15

1.165

0.05

1.171509

1.241

1.310991

0

0.096

0.218311

1.2

1.21

1.2

1.21

1.205

0.01

1.171509

1.241

1.310991

0

0.096

0.218311

1.25

1.17

1.2

1.24

1.215

0.08

1.171509

1.241

1.310991

0

0.096

0.218311

1.26

1.25

1.3

1.22

1.258

0.08

1.171509

1.241

1.310991

0

0.096

0.218311

1.24

1.19

1.22

1.21

1.215

0.05

1.171509

1.241

1.310991

0

0.096

0.218311

1.36

1.18

1.36

1.16

1.265

0.20

1.171509

1.241

1.310991

0

0.096

0.218311

1.25

1.15

1.29

1.15

1.210

0.14

1.171509

1.241

1.310991

0

0.096

0.218311

1.3

1.32

1.29

1.23

1.285

0.09

1.171509

1.241

1.310991

0

0.096

0.218311

1.29

1.34

1.2

1.37

1.300

0.17

1.171509

1.241

1.310991

0

0.096

0.218311

1.2

1.24

1.14

1.23

1.203

0.10

1.171509

1.241

1.310991

0

0.096

0.218311

1.26

1.28

1.25

1.29

1.270

0.04

1.171509

1.241

1.310991

0

0.096

0.218311

1.17

1.3

1.15

1.3

1.230

0.15

1.171509

1.241

1.310991

0

0.096

0.218311

1.28

1.25

1.21

1.26

1.250

0.07

1.171509

1.241

1.310991

0

0.096

0.218311

1.19

1.16

1.25

1.27

1.218

0.11

1.171509

1.241

1.310991

0

0.096

0.218311

1.23

1.24

1.23

1.22

1.230

0.02

1.171509

1.241

1.310991

0

0.096

0.218311

1.22

1.23

1.26

1.23

1.235

0.04

1.171509

1.241

1.310991

0

0.096

0.218311

1.24

1.25

1.16

1.2

1.213

0.09

1.171509

1.241

1.310991

0

0.096

0.218311

1.32

1.19

1.25

1.19

1.238

0.13

1.171509

1.241

1.310991

0

0.096

0.218311

1.25

1.21

1.21

1.22

1.223

0.04

1.171509

1.241

1.310991

0

0.096

0.218311

1.17

1.17

1.29

1.29

1.230

0.12

1.171509

1.241

1.310991

0

0.096

0.218311

1.14

1.14

1.35

1.24

1.218

0.21

1.171509

1.241

1.310991

0

0.096

0.218311

1.241

0.096

XDblBar

R-bar

(ii) Does the process appear to be in statistical control? What evidence is there to support your answer? (5 Marks)

No points are out of control on the R chart. Two-point are out of control on the Xbar chart which indicates a high variation range in the process. Therefore, the process is not in statistical control.







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