(Solution Manual) Quality Control

Montgomery, 5

Math & Statistics

Solution Manual

5th Edition

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Solution Manual for Quality Control, The downloadable file consists of 472 pages, and the total size of the file is 6.96 MB. Purchase now only for highly discounted price.

Author(s): Montgomery

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Chapter 2 Exercise Solutions Several exer cises in th is chapte r differ from those in the 4th edition. An “*” following the exercise number indicates that the descrip tion has changed (e.g., new values). A second exercise number in parenthe ses indicates that th e exerc ise num ber has changed. For exam ple, “2-16* (2-9)” m eans that exercise 2-16 was 2-9 in the 4th editio n, and that the description also differs from the 4th edition (in this case, ask ing for a tim e series plo t instead of a digidot plot). New exercises are den oted with an “☺”. 2-1*. (a) () 116.05 16.03 16.07 12 16.029 ozn i ixx n === +++ = ∑ " (b) ()2 2 22 2 11 (16.05 16.07 )(16.05 16.07 )120.0202 oz11 21nn ii iixx n sn==−∑∑++ −++== =−−"" MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-1 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-1 12 0 16.029 0.00583 0.0202 16.000 16.013 16.025 16.048 Variable Maximum Ex2-1 16.070 2-2. (a) () 150.001 49.998 50.004 850.002 mmn i ixx n === +++ = ∑ " (b) ()2 2 22 2 11 (50.001 50.004 )(50.001 50.004) 80.003 m m18 1nn ii iixx n sn==−∑∑++ −++== =−−"" MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-2 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-2 8 0 50.002 0.00122 0.00344 49.996 49.999 50.003 50.005 Variable Maximum Ex2-2 50.006 2-1Chapter 2 Exercise Solutions 2-3. (a) () 1953 955 959 9952.9 °Fn i ixx n === +++= ∑ " (b) ()2 2 22 2 11 (953 959)(953 959) 93.7 °F19 1nn ii iixx n sn==−∑∑++ −++==−−""= MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-3 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-3 9 0 952.89 1.24 3.72 948.00 949.50 953.00 956.00 Variable Maximum Ex2-3 959.00 2-4. (a) In ranked order, the data are {948, 949, 950, 951, 953, 954, 955, 957, 959}. The sample median is the m iddle value. (b) Since the m edian is the value dividing the ra nked sam ple observations in half, it rem ains the sam e regardles s of the size of the larges t measurem ent. 2-5. MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-5 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-5 8 0 121.25 8.00 22.63 96.00 102.50 117.00 144.50 Variable Maximum Ex2-5 156.00 2-2Chapter 2 Exercise Solutions 2-6. (a), (d ) MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-6 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-6 40 0 129.98 1.41 8.91 118.00 124.00 128.00 135.25 Variable Maximum Ex2-6 160.00 (b) Use √n = √40 ≅ 7 bins MTB > Graph > Histogram > Simple Hours 160 152 144 136 128 120 11220 15 10 5 0Histogram of Time to Failure (Ex2-6) (c) MTB > Graph > Stem-and-Leaf Stem-and-Leaf Displa y: Ex2-6 Stem-and-leaf of Ex2-6 N = 40 Leaf Unit = 1.0 2 11 89 5 12 011 8 12 233 17 12 444455555 19 12 67 (5) 12 88999 16 13 0111 12 13 33 10 13 10 13 677 7 13 7 14 001 4 14 22 HI 151, 160 2-3Chapter 2 Exercise Solutions 2-7. Use 90 9 n=≅ bins MTB > Graph > Histogram > Simple Yield 96 92 88 8418 16 14 12 10 8 6 4 2 0Histogram of Process Yield (Ex2-7) 2-4Chapter 2 Exercise Solutions 2-8. (a) Stem-and-Leaf Plot 2 12o|68 6 13*|3134 12 13o|776978 28 14*|3133101332423404 (15) 14o|585669589889695 37 15*|3324223422112232 21 15o|568987666 12 16*|144011 6 16o|85996 1 17*|0 Stem Freq|Leaf (b) Use 80 9 n=≅ bins MTB > Graph > Histogram > Simple Viscosity 17 16 15 14 1320 15 10 5 0Histogram of Viscosity Data (Ex 2-8) Note that the histogram has 10 bins. The num ber of bins can be changed by editing the X scale. However, if 9 bins are specif ied, MINITAB generates an 8-bin histogram . Constructing a 9-bin histogram requires m anual specification of the bin cut points. Recall that this form ula is an approxim ation, and therefore either 8 or 10 bins should suffice for assessing the dist ribution of the data. 2-5Chapter 2 Exercise Solutions 2-8(c) continued MTB > %hbins 12.5 17 .5 c7 Row Intervals Frequencies Percents 1 12.25 to 12.75 1 1.25 2 12.75 to 13.25 2 2.50 3 13.25 to 13.75 7 8.75 4 13.75 to 14.25 9 11.25 5 14.25 to 14.75 16 20.00 6 14.75 to 15.25 18 22.50 7 15.25 to 15.75 12 15.00 8 15.75 to 16.25 7 8.75 9 16.25 to 16.75 4 5.00 10 16.75 to 17.25 4 5.00 11 Totals 80 100.00 (d) MTB > Graph > Stem-and-Leaf Stem-and-Leaf Displa y: Ex2-8 Stem-and-leaf of Ex2-8 N = 80 Leaf Unit = 0.10 2 12 68 6 13 1334 12 13 677789 28 14 0011122333333444 (15) 14 555566688889999 37 15 1122222222333344 21 15 566667889 12 16 011144 6 16 56899 1 17 0 median observation rank is (0.5)(80) + 0.5 = 40.5 x0.50 = (14.9 + 14.9)/2 = 14.9 Q1 observation rank is (0.25)(80) + 0.5 = 20.5 Q1 = (14.3 + 14.3)/2 = 14.3 Q3 observation rank is (0.75)(80) + 0.5 = 60.5 Q3 = (15.6 + 15.5)/2 = 15.55 (d) 10th percentile observation ra nk = (0.10)(80) + 0.5 = 8.5 x0.10 = (13.7 + 13.7)/2 = 13.7 90th percentile observation ra nk is (0.90)(80) + 0.5 = 72.5 x0.90 = (16.4 + 16.1)/2 = 16.25 2-6Chapter 2 Exercise Solutions 2-9 ☺. MTB > Graph > Probability Plot > Single Fluid Ounces 16.08 16.06 16.04 16.02 16.00 15.9899 95 90 80 70 60 50 40 30 20 10 5 1Mean 0.53216.03 StDev 0.02021 N1 AD 0.297 P-ValueNormal Probability Plot of Liquid Detergent (Ex2-1) 2 When plotted on a normal probability plot, th e data points tend to fall along a straight line, indicating that a no rmal distribution adequately descri bes the volum e of detergent. 2-10 ☺. MTB > Graph > Probability Plot > Single Temperature (deg F) 962.5 960.0 957.5 955.0 952.5 950.0 947.5 945.099 95 90 80 70 60 50 40 30 20 10 5 1Mean 0.908952.9 StDev 3.723 N9 AD 0.166 P-ValueNormal Probability Plot of Furnace Temperatures (Ex2-3) When plotted on a normal probability plot, th e data points tend to fall along a straight line, indicating that a no rmal distribution ad equately describes the furnace tem peratures . 2-7Chapter 2 Exercise Solutions 2-11 ☺. MTB > Graph > Probability Plot > Single Hours 160 150 140 130 120 11099 95 90 80 70 60 50 40 30 20 10 5 1Mean <0.005130.0 StDev 8.914 N4 AD 1.259 P-ValueNormal Probability Plot of Failure Times (Ex2-6) 0 When plotted on a normal probability plot, the da ta points do not fall al ong a straight line, indicating that the norm al distribution does not reasonably de scrib e the f ailure tim es. 2-12 ☺. MTB > Graph > Probability Plot > Single Yield 105 100 95 90 85 8099.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1Mean 0.01589.48 StDev 4.158 N9 0 AD 0.956 P-ValueNormal Probability Plot of Process Yield Data (Ex2-7) When plotted on a normal probability plot, the da ta points do not fall al ong a straight line, indicating that the norm al distribution does not reasonabl y describe process yield. 2-8Chapter 2 Exercise Solutions 2-13 ☺. MTB > Graph > Probability Plot > Single (In the dialog box, select Distribu tion to choose th e distributions) Viscosity 18 17 16 15 14 13 1299.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1Mean 0.74014.90 StDev 0.9804 N8 AD 0.249 P-ValueNormal Probability Plot of Viscosity Data (Ex2-8) 0 Viscosity 19 18 17 16 15 14 13 1299.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1Loc 0.8412.699 Scale 0.06595 N8 AD 0.216 P-ValueLognormal Probability Plot of Viscosity Data (Ex2-8) 0 2-9Chapter 2 Exercise Solutions 2-13 continued Viscosity 18 17 16 15 14 13 12 11 1099.9 99 90 80 70 60 50 40 30 20 10 5 3 2 1 0.1Shape <0.01016.10 Scale 15.36 N8 AD 1.032 P-ValueWeibull Probability Plot of Viscosity Data (Ex2-8) 0 Both the norm al and lognorm al distributions app ear to be reasonable m odels for the data; the plot points tend to fall al ong a straight line, w ith no bends or curves. However, the plot points on the W eibull probability plot are not straight—particula rly in the ta ils— indicating it is not a reasonable m odel. 2-10Chapter 2 Exercise Solutions 2-14 ☺. MTB > Graph > Probability Plot > Single (In the dialog box, select Distribu tion to choose th e distributions) Cycles to Failure 25000 20000 15000 10000 5000 0 -500099 95 90 80 70 60 50 40 30 20 10 5 1Mean 0.1378700 StDev 6157 N2 0 AD 0.549 P-ValueNormal Probability Plot of Cycles to Failure (Ex2-14) Cycles to Failure 100000 10000 100099 95 90 80 70 60 50 40 30 20 10 5 1Loc 0.1638.776 Scale 0.8537 N2 AD 0.521 P-ValueLognormal Probability Plot of Cycles to Failure (Ex2-14) 0 2-11Chapter 2 Exercise Solutions 2-14 continued Cycles to Failure 10000 100099 90 80 70 60 50 40 30 20 10 5 3 2 1Shape >0.2501.464 Scale 9624 N2 AD 0.336 P-ValueWeibull Probability Plot of Cycles to Failure (Ex2-14) 0 Plotted points do not tend to fall on a straight line on any of the probability plots, though the W eibull distribution appears to best fit the da ta in the ta ils. 2-12Chapter 2 Exercise Solutions 2-15 ☺. MTB > Graph > Probability Plot > Single (In the dialog box, select Distribu tion to choose th e distributions) Concentration, ppm 150 100 50 0 -5099 95 90 80 70 60 50 40 30 20 10 5 1Mean <0.0059.470 StDev 22.56 N4 AD 8.426 P-ValueNormal Probability Plot of Concentration (Ex2-15) 0 Concentration, ppm 100.0 10.0 1.0 0.199 95 90 80 70 60 50 40 30 20 10 5 1Loc 0.8730.9347 Scale 1.651 N4 AD 0.201 P-ValueLognormal Probability Plot of Concentration (Ex2-15) 0 2-13Chapter 2 Exercise Solutions 2-15 continued Concentration, ppm 100.000 10.000 1.000 0.100 0.010 0.00199 90 80 70 60 50 40 30 20 10 5 3 2 1Shape 0.0910.6132 Scale 5.782 N4 AD 0.637 P-ValueWeibull Probability Plot of Concentration (Ex2-15) 0 The lognorm al distribution appears to be a reas onable m odel for the concentration data. Plotted po ints on the no rmal and W eibull prob ability p lots ten d to fall off a stra ight lin e. 2-14Chapter 2 Exercise Solutions 2-16* (2-9). MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Order of Collection 80 72 64 56 48 40 32 24 16 8 117 16 15 14 13 12Time Series Plot of Viscosity Data (Ex2-8) From visual exam ination, there are no trends , shifts or obvious pa tterns in the data, indicating that tim e is not an im portant source of variability. 2-17* (2-10). MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Order of Collection 90 81 72 63 54 45 36 27 18 9 1100 95 90 85Time Series Plot of Yield Data (Ex2-7) Time may be an im portant source of variab ility, as evidenced by potentially cyclic behavior. 2-15Chapter 2 Exercise Solutions 2-18 ☺. MTB > Graph > Time Series Plot > Single (or Stat > Time Series > Time Series Plot) Time Order of Collection 40 36 32 28 24 20 16 12 8 4140 120 100 80 60 40 20 0Time Series Plot of Concentration Data (Ex2-15) Although most of the readings are between 0 and 20, there are two unusually large readings (9, 35), as well as occasional “spi kes” around 20. The order in which the d ata were collected m ay be a n important source of variability. 2-19 (2-11). MTB > Stat > Basic Statistics > Displa y Desc riptive Statistics Descriptive Statistics: Ex2-7 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Ex2-7 90 0 89.476 0.438 4.158 82.600 86.100 89.250 93.125 Variable Maximum Ex2-7 98.000 2-16Chapter 2 Exercise Solutions 2-20 (2-12). MTB > Graph > Stem-and-Leaf Stem-and-Leaf Displa y: Ex2-7 Stem-and-leaf of Ex2-7 N = 90 Leaf Unit = 0.10 2 82 69 6 83 0167 14 84 01112569 20 85 011144 30 86 1114444667 38 87 33335667 43 88 22368 (6) 89 114667 41 90 0011345666 31 91 1247 27 92 144 24 93 11227 19 94 11133467 11 95 1236 7 96 1348 3 97 38 1 98 0 Neither the stem -and-leaf pl ot nor the frequency histogr am reveals m uch about an underlying distribution or a central tendency in the data. The data appear to be fairly w ell scattered. T he stem -and-leaf plot suggests that certain values m ay occur more frequently than others; for exam ple, those ending in 1, 4, 6, and 7. 2-21 (2-13). MTB > Graph > Boxplot > Simple 16.07 16.06 16.05 16.04 16.03 16.02 16.01 16.00Boxplot of Detergent Data (Ex2-1) 2-17Chapter 2 Exercise Solutions 2-22 (2-14). MTB > Graph > Boxplot > Simple 50.0075 50.0050 50.0025 50.0000 49.9975 49.9950Boxplot of Bearing Bore Diameters (Ex2-2) 2-23 (2-15). x: {the sum of two up dice faces} sample space : {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 11 1 Pr{ 2}Pr{1,1}66 3x== =×=6 ()()11 11 2 Pr{ 3}Pr{1,2}Pr{2,1}66 66 3x== +=×+×=6 ()()()3 11 11 11 Pr{ 4}Pr{1,3}Pr{2,2}Pr{3,1}66 66 66 3x==++=×+× +× =6 . . . 1/36; 22/36; 33/36; 44/36; 55/36; 66/36; 7()5/36;84/36; 93/36;102/36;111/36;120; other wisexx xx x xpxxx x x x== == = = ⎧=⎨== = = = ⎩ 2-24 (2-16). ()() ()11 1()2136 3236 121367ii ixx px === +++ ∑ " = 2 2 11() ()5.92 7110.3811 0nn ii i i iixpx xpx n Sn==⎡⎤−∑∑⎢⎥ − ⎣⎦==−= 2-18Chapter 2 Exercise Solutions 2-25 (2-17). This is a Po isson distribu tion with p arameter λ = 0.02, x ~ POI(0.02). (a) 0.02 1(0.02)Pr{1} (1) 0.01961!exp− == = = (b) 0.02 0(0.02)Pr{1}1Pr{ 0}1 (0)1 10.9802 0.01980!exx p− ≥=−==−=− =− = (c) This is a Po isson distribu tion with p arameter λ = 0.01, x ~ POI(0.01). 0.01 0(0.01)Pr{1}1Pr{ 0}1 (0)1 10.9900 0.01000!exx p− ≥=−==−=− =− = Cutting th e rate at which def ects occu r reduc es the probability of one or m ore def ects b y approxim ately one-half, from 0.0198 to 0.0100. 2-26 (2-18). For f(x) to be a probability distribution, ()fxdx+∞ −∞∫ must equal unity. 0 0[] [01]xxkedx ke k k∞−− ∞=− =−−=⇒ ∫ 1 = This is an exponential di stribution with param eter λ=1. µ = 1/λ = 1 (Eqn. 2-32) σ2 = 1/λ2 = 1 (Eqn. 2-33) 2-27 (2-19). (13)/3; 1(12)/3; 2()(0.55)/3; 30; othe rwisekx kxpxkx+= + ⎧=⎨+=⎩ (a) To solve for k, use 1() ()1i iFx px∞ ===∑ (13)(12)(0.55)13 10 0.5 0.05kk k k k+++++= = = 2-19Chapter 2 Exercise Solutions 2-27 continued (b) 3 113(0.05) 12(0.05) 0.55(0.05)()1 2 3 1.86733 3ii ixpx µ =++ + ⎡⎤ ⎡ ⎤⎡ ⎤== × +× +× = ∑⎢⎥ ⎢ ⎥⎢ ⎥⎣⎦ ⎣ ⎦⎣ ⎦ 322 2 2 2 2 2 1() 1(0.383 )2(0.367)3(0.250) 1.867 0.615ii ixp x σµ ==− = + + −= ∑ (c) 1.150.383 ;13 1.15 1.1() 0.750 ; 23 1.15 1.1 0.751.000 ;33x Fx x x⎧==⎪ ⎪+ ⎪==⎨ ⎪ ++⎪==⎪⎩= 2-28 (2-20). ()0() ; 0 1; 0,1,2, () 1 by definition 11 1 1x x ipx kr r x Fx kr kr kr∞ ==< <= ==∑ ⎡⎤−=⎣⎦ =−… 2-29 (2-21). (a) This is an exponential di stribution with param eter λ = 0.125: 0.125(1)Pr{1} (1)1 0.118 xF e−≤= =−= Approxim ately 11.8% w ill fail dur ing the f irst year. (b) Mfg. cost = $50/calculator Sale profit = $25/calculator Net profit = $[-50(1 + 0.118) + 75] /calculator = $19.10/calculator. The effect of warranty replacem ents is to decrease profit by $5.90/calculator. 2-20Chapter 2 Exercise Solutions 2-30 (2-22). 1212 12 2 12 11.75 11.75 11.754Pr{ 12} (12) () 4(11.75 ) 47 11.87511.75 0.125 2xxF fxdx x dx x −∞<= = =− =−=−= ∫∫ 2-31* (2-23). This is a b inomial dis tribution with p arameter p = 0.01 and n = 25. The process is stopped if x ≥ 1. 02 5 25Pr{1}1Pr{ 1}1Pr{ 0}1 (0.01)(10.01) 10.78 0.220xxx⎛⎞≥=−<=−==− −=−= ⎜⎟⎝⎠ This decisio n rule m eans that 22 % of the sam ples will have o ne or m ore nonconfor ming units, and the process will be stopped to look for a cause. This is a som ewhat difficult operating situation. This exercis e may also be solved us ing Excel or MINITAB: (1) Excel F unction BINOMDIST(x , n, p, TRUE) (2) MTB > Calc > Probability Distri butions > Binomial Cumulative Distribution Function Binomial with n = 25 and p = 0.01 x P( X <= x ) 0 0.777821 2-32* (2-24). x ~ BIN(25, 0.04) Stop process if x ≥ 1. 02 5 25Pr{1}1Pr{ 1}1Pr{ 0}1 (0.04) (10.04) 10.36 0.640xxx⎛⎞≥=−<=−==− −=−= ⎜⎟⎝⎠ 2-33* (2-25). This is a b inomial dis tribution with p arameter p = 0.02 and n = 50. 4 (50) 0 05 0 1 49 4 4650ˆPr{ 0.04}Pr{ 2} (0.02) (10.02) 50 50 50(0.02) (10.02) (0.02) (10.02) (0.02) (10.02) 0.92101 4xx xpxx− =⎛⎞≤= ≤= − ⎜⎟⎝⎠ ⎛⎞ ⎛⎞ ⎛⎞=− + −++ −= ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠∑ " 2-21Chapter 2 Exercise Solutions 2-34* (2-26). This is a b inomial dis tribution with p arameter p = 0.01 and n = 100. 0.01(10.01)100 0.0100 σ=− = ˆˆPr{ }1Pr{ }1Pr{ ( )} pkp pk p xnk p σ σσ >+=− ≤+=−≤+ k = 1 2100 0 01 00 19 9 2 91Pr{ ( )}1Pr{ 100(1(0.0100) 0.01)}1Pr{ 2} 1001 (0.01)(10.01) 100 100 1001 (0.01)(0.99) (0.01)(0.99) (0.01)(0.99)01 2 1[0.921 ]0.079xx xxn k p x x x 8σ − =−≤ +=−≤ +=−≤ ⎛⎞=− − ∑⎜⎟⎝⎠ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞=− + + ⎢⎥⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎣⎦ =− = k = 2 3100 3 97 01Pr{ ( )}1Pr{ 100(2(0.0100) 0.01)}1Pr{ 3} 100 1001 (0.01)(0.99) 10.921 (0.01)(0.99)3 1[0.982] 0.018xx xxn k p x x xσ − =−≤ +=−≤ +=−≤ ⎡ ⎤ ⎛⎞ ⎛⎞=− =− + ∑ ⎢ ⎥ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠⎣ ⎦ =− = k = 3 4100 4 96 01Pr{ ( )}1Pr{ 100(3(0.0100) 0.01)}1Pr{ 4} 100 1001 (0.01)(0.99) 10.982 (0.01)(0.99)4 1[0.992] 0.003xx xxn k p x x xσ − =−≤ +=−≤ +=−≤ ⎡ ⎤ ⎛⎞ ⎛⎞=− =− + ∑ ⎢ ⎥ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠⎣ ⎦ =− = 2-22Chapter 2 Exercise Solutions 2-35* (2-27). This is a hy pergeom etric distribu tion with N = 25 and n = 5, without replacem ent. (a) Given D = 2 and x = 0: 22 52 05 0 (1)(33,649)Pr{Acceptance} (0) 0.63325 (53,130) 5p−⎛⎞⎛⎞ ⎜⎟⎜⎟−⎝⎠⎝⎠== = =⎛⎞ ⎜⎟⎝⎠ This exercis e may also be solved us ing Excel or MINITAB: (1) Excel F unction HYPGEOMDIST(x, n, D, N) (2) MTB > Calc > Probability Distri butions > Hy pergeome tric Cumulative Distribution Function Hypergeometric with N = 25, M = 2, and n = 5 x P( X <= x ) 0 0.633333 (b) For the binom ial approxim ation to the hypergeom etric, p = D/N = 2/25 = 0.08 and n = 5. 0 5 5Pr{acceptanc e} (0) (0.08)(10.08)0.6590p⎛⎞== −= ⎜⎟⎝⎠ This approxim ation, though clos e to the exact solution for x = 0, violates the rule-of- thum b that n/N = 5/25 = 0.20 be less than the sugges ted 0.1. The binom ial approxim ation is not sa tisfactory in this case. (c) For N = 150, n/N = 5/150 = 0.033 ≤ 0.1, so the binom ial approxim ation would be a satisfactory approxim ation the hypergeom etric in this case. 2-23Chapter 2 Exercise Solutions 2-35 continued (d) Find n to satisfy Pr{ x ≥ 1 | D ≥ 5} ≥ 0.95, or equivalently Pr{ x = 0 | D = 5} < 0.05. 52 55 52 0 00 0(0)25 25nnp nn−⎛⎞⎛⎞⎛⎞⎛⎞ ⎜⎟⎜⎟⎜⎟⎜⎟−⎝⎠⎝⎠⎝⎠⎝⎠==⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ try 1 0 52 0 01 0 (1)(184,756)(0) 0.05725 (3,268,760) 10 try 1 1 52 0 01 1 (1)(167,960)(0) 0.03825 (4,457,400) 11n p n p= ⎛⎞⎛⎞ ⎜⎟⎜⎟⎝⎠⎝⎠== =⎛⎞ ⎜⎟⎝⎠ = ⎛⎞⎛⎞ ⎜⎟⎜⎟⎝⎠⎝⎠== =⎛⎞ ⎜⎟⎝⎠ Let sam ple size n = 11. 2-36 (2-28). This is a hy pergeom etric distribu tion with N = 30, n = 5, and D = 3. 33 03 15 1 (3)(17,550)Pr{1} (1) 0.36930 (142,506) 5xp−⎛⎞⎛⎞ ⎜⎟⎜⎟−⎝⎠⎝⎠== = = =⎛⎞ ⎜⎟⎝⎠ 32 7 05Pr{1}1Pr{ 0}1 (0)1 10.567 0.43330 5xx p⎛⎞⎛⎞ ⎜⎟⎜⎟⎝⎠⎝⎠≥=−==−=− =− =⎛⎞ ⎜⎟⎝⎠ 2-24Chapter 2 Exercise Solutions 2-37 (2-29). This is a hy pergeom etric distribu tion with N = 500 pages, n = 50 pages, and D = 10 errors. Checking n/N = 50/500 = 0.1 ≤ 0.1, the binom ial distribution can be used to approxim ate the hypergeom etric, with p = D/N = 10/500 = 0.020. 05 00 50Pr{ 0} (0) (0.020) (10.020) (1)(1)(0.364) 0.3640xp− ⎛⎞== = − = = ⎜⎟⎝⎠ 15 01Pr{ 2}1Pr{1}1[Pr{ 0}Pr{1}]1 (0)(1) 5010.364 (0.020) (10.020) 10.364 0.372 0.2641xx x x p p −≥=−≤=−=+==−− ⎛⎞=− − − =− −= ⎜⎟⎝⎠ 2-38 (2-30). This is a Poisson distribution with λ = 0.1 defects/unit. 0.1 0(0.1)Pr{1}1Pr{ 0}1 (0)1 10.905 0.0950!exx p− ≥=−==−=− =− = This exercis e may also be solved us ing Excel or MINITAB: (1) Excel F unction POISSON( λ, x, TRUE) (2) MTB > Calc > Probability Distri butions > Poisson Cumulative Distribution Function Poisson with mean = 0.1 x P( X <= x ) 0 0.904837 2-39 (2-31). This is a Poisson distribution with λ = 0.00001 stones/bottle. 0.00001 0(0.00001 )Pr{1}1Pr{ 0}1 10.99999 0.000010!exx− ≥=−==− =− = 2-40 (2-32). This is a Poisson distribution with λ = 0.01 erro rs/bill. 0.01 1(0.01)Pr{1} (1) 0.00991exp− == = = 2-25