Chapter 10 Solutions - California State University, …



Chapter 10 Solutions

1. specs: 24 oz. to 25 oz.

( = 24.5 oz. [assume ( = ]

( = .2 oz.

a. [refers to population]

[pic]

b. [pic]

2. ( = 1.0 liter

( = .01 liter

n = 25

a. [pic]

[z = 2.17 for 97%]

[pic]

|3. a. |n = 20 |

| | |A2 = 0.18 | = 3.10 Mean Chart: ( A2 = 3.1 ( 0.18(0.45) |

| | |D3 = 0.41 |= 0.45 |= 3.1 ( .081 |

| | |D4 = 1.59 | |Hence, UCL is 3.181 |

| | | | |and LCL is 3.019. All means are within these limits. |

| | | | |Range Chart: UCL is D4= 1.59(0.45) = .7155 |

| | | | |LCL is D3= 0.41(0.45) = .1845 |

| | | | |In control since all points are within these limits. |

|4. |Sample |Mean |Range | |

| |1 |79.48 |2.6 |Mean Chart: ( A2= 79.96 ( 0.58(1.87) |

| |2 |80.14 |2.3 | |= 79.96 ( 1.1 |

| |3 |80.14 |1.2 | |UCL = 81.04, LCL = 78.88 |

| |4 |79.60 |1.7 |Range Chart: UCL = D4= 2.11(1.87) = 3.95 |

| |5 |80.02 |2.0 | |LCL = D3= 0(1.87) = 0 |

| |6 |80.38 |1.4 |[Both charts suggest the process is in control: Neither has any points outside the |

| | | | |limits.] |

Solutions (continued)

5. n = 200

| |a. |1 |2 |3 |4 |

| | |.020 |.010 |.025 |.045 |

b. (2.0 + 1.0 + 2.5 + 4.5)/4 = 2.5%

c. mean = .025

[pic]

d. z = 2.17

.025 ( 2.17(0.011) = .025% ( .0239% = .0011 to .0489.

e. .025 + z(.011) = .047

Solving, z = 2, leaving .0228 in each tail. Hence, alpha = 2(.0228)

= .0456.

f. Yes.

g. mean = .02

[pic]

h. .02 ( 2(.01) = 0 to .04

The last sample is beyond the upper limit.

6. n = 200 Control Limits = [pic]

Thus, UCL is .0234 and LCL becomes 0.

Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.

Sample 10 is too large. Omitting that value and recomputing limits with

[pic]

UCL = .0197 and LCL = 0.

7. [pic] Control limits: [pic]

UCL is 16.266, LCL becomes 0.

All values are within the limits.

Solutions (continued)

8. [pic] Control limits: [pic]

UCL is 5.17, LCL becomes 0.

All values are within the limits.

9.

[pic]

[pic] Hence, UCL = 0.10

LCL = 0.01

Note that observations must be converted to fraction defective, or control limits must be converted to number of defectives. In the latter case, the upper control limit would be 7.9 defectives and the lower control limit would be .1 defective. Even though all points are within these limits, the process appears to be out of control because 75% of the values are above 4%.

10. There are several slightly different ways to solve this problem. The most straightforward seems to be the following:

1) Observe that the upper control limit is six standard deviations above the lower control limit.

2) Compute the value of the upper control limit at the start:

[pic]

3) Determine how many pieces can be produced before the upper control limit just touches the upper tolerance, given that the upper limit increases by .004 cm. per piece:

| |15.2cm. – 15.06cm. |= 35 pieces. |

| |.004 cm./piece | |

11. Out of the 30 observations, only one value exceeds the tolerances, or 3.3%. [This case is essentially the one portrayed in the text in Figure 10–9A.] Thus, it seems that the tolerances are being met: approximately 97 percent of the output will be acceptable.

12. a. ( = .146

n = 14

[pic]

Control limits are [pic]

So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the process is not in control.

Solutions (continued)

b. [median is 3.85]

| | |Sample |A/B |Mean |U/D | |Sample |A/B |Mean |U/D |

| | |1 |A |3.86 |( | |21 |

| | |Median |18 |20.5 |3.08 |–.81 |random |

| | |Up/down |29 |25.7 |2.57 |1.28 |random |

13. a.

| |A A B B A B A B B A B A A B A A A B B B A B A B A B |

| | |

b.

| |

| |A A A A B A A B B B B B A B B B A A A A B B B B B B |

| | |

Solutions (continued)

| |Summary: |obs. |exp. |( |z |Conclusion |

| |a. |median |18 |14 |2.50 |1.6 |random |

| |up/down |17 |17 |2.07 |0.0 |random |

| |b. |median |8 |14 |2.50 |–2.40 |nonrandom |

| |up/down |22 |17 |2.07 |2.41 |nonrandom |

14. a. [Data from Chapter 10, Problem 8]

Median is 1.5 A = Above, B = Below, U = Up, D = Down.

| | |Sample: |1 |2 |3 |4 |5 |6 |7 |

| |1 |B |

(ii) When UCL = 3.5 Cm, the LCL = 3.5 – 6 [pic]

| |(iii) |Determine how many pieces can be produced before the LCL just crosses the lower tolerance of 3 Cm. |

| | | |3.44 – 3.00 |= |0.44 |= |440 |= 440 pieces |

| | | |0.001 | |0.001 | |1 | |

17. It is necessary to see if the process variability is within 9.96 and 10.35. Two observations have values above the specified limits, i.e., 10% of the 20 observations fall outside the limits. Perhaps the process mean should be set a bit lower.

18. 1 Step 10% scrap, 2nd 6%, and 3rd 6%.

a. Let x be the number of units started initially at Step 1. With a scrap rate of 10% in Step 1 the input to Step 2 is 0.9x. The input to Step 3 is (1 – 0.06) (1 – 0.10)x. With a scrap rate of 6% at Step 3 the number of good units after Step 3 = (1 – 0.06) (1 – 0.06) (1 – 0.10)x = (0.94)2(0.90)x.

The required output is 450 units

(0.94)2(0.90)x = 450 units

x = 565.87 ( 566 units

b. (1 – 0.03)2(1 – 0.05)x = 450 units

(0.97)2(0.95)x = 450 units

x = 503.44 ( 504 units

Savings of 566 – 504 = 62 units

c. From (a)

The scrap = 566 – 450 = 116 units

@ $10 per unit, The Total Cost = $1,160.00

Solutions (continued)

19. Sample #

| | | |

| | |Observed | |Expected | |Deviation | |Z | |Conclude |

| |Median |13 | |11 | |2.1794 | |0.9177 | |Random |

| |Up/down |14 | |13 | |1.7981 | |0.5561 | |Random |

|20. |a. |1 |2 |3 |4 |

| | |4.3 |4.5 |4.5 |4.7 |

b. = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5

std. dev. (of data set) = .192

c. mean = 4.5, std. dev. = [pic]

d. 4.5 ( 3(.086) = 4.5 ( .258 = 4.242 to 4.758

The risk is 2(.0013) = .0026

e. 4.5 + z(.086) = 4.86

Solving, z = 4.19, so the risk is close to zero

f. None

g. = (.3 + .4 + .2 + .4)/4 = .325

n = 5

Means: A2 = 0.58 ( A2= 4.5 ( 0.58(.325) = 4.3115 to 4.6885.

The last mean is above the upper limit.

Ranges: D3 = 0 0 to 2.11(.325) = 0 to .6875 All ranges are within the limits.

h. Two different measures of dispersion are being used, the standard deviation and the range.

i. [pic]to 4.64. The last value is above the upper limit.

21. Solution

a. [pic]

b. In order to be capable, the process capability ratio must be at least 1.33. In this instance, the index is 1.11, so the process is not capable.

Solutions (continued)

|22. |Machine |Standard Deviation (in.) |Job Specification ((in.) |Cp |Capable ? |

| |001 |0.02 |0.05 |0.833 |No |

| |002 |0.04 |0.07 |0.583 |No |

| |003 |0.10 |0.18 |0.600 |No |

| |004 |0.05 |0.15 |1.000 |No |

| |005 |0.01 |0.04 |1.333 |Yes |

|23. |Machine |Cost per unit ($) |Standard Deviation (mm.) |Cp |

| |A |20 |0.059 |1.355 |

| |B |12 |0.060 |1.333 |

| |C |11 |0.063 |1.27 |

| |D |10 |0.061 |1.311 |

You can narrow the choice to machines A and B because they are the only ones with a capability ratio of at least 1.33. You would need to know if the slight additional capability of machine A is worth an extra cost of $8 per unit.

24. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,

[pic]= Process mean, ( = Process standard deviation

For process H:

[pic]

For process K:

[pic]

Since 1.0 < 1.33, the process is not capable.

Solutions (continued)

For process T:

[pic]

Since 1.33 = 1.33, the process is capable.

25. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,

[pic]= Process mean, ( = Process standard deviation.

USL = 90 minutes, LSL = 50 minutes,

[pic] = 74 minutes, [pic]= 4.0 minutes

[pic] = 72 minutes, [pic]= 5.1 minutes

For the first repair firm:

[pic]

Since 1.333 = 1.333, the firm 1 is capable.

For the second repair firm:

[pic]

Since 1.18 < 1.33, the firm 2 is not capable.

Solutions (continued)

26. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,

[pic]= Process mean, ( = Process standard deviation.

USL = 30 minutes, LSL = 45 minutes,

[pic]Armand = 38 minutes, ( Armand = 3 minutes

[pic]Jerry = 37 minutes, ( Jerry = 2.5 minutes

[pic]Melissa = 37.5 minutes, ( Melissa = 2.5 minutes

For Armand:

[pic]

Since .78 < 1.33, Armand is not capable.

For Jerry:

[pic]

Since .93 < 1.33, Jerry is not capable.

For Melissa, since[pic], the process is centered, therefore we will use Cp to measure process capability.

[pic]

Since 1.39 > 1.33, Melissa is capable.

-----------------------

.0062

.0062

24 24.5 25 16

-2.5 0 +2.5 z-scale

1.006

1.0043

1.002

1.000

.998

.9957

.994

UCL

LCL

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out

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Mean

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3.5

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3.0 Cm

3 sigma

n = 1

( = 0.01 cm

UCL

Mean

LCL

3 sigma

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