《质量管理学》课程综合练习题

34. 解：（计算表）

组号 1 2 3 4 5 6 7 8 9 10 X1 40.3 39.95 39.88 40.02 39.92 40.09 39.95 39.85 40.08 39.98 X2 40.02 39.92 40.24 39.96 40.07 39.94 40.06 40.03 39.95 40 观测值 X3 40.19 40.01 40.21 39.93 40.15 39.97 39.94 39.93 40.09 39.9 X4 39.92 40.01 40.05 40.15 39.89 39.85 40 40.15 40.05 40.07 X5 40.08 40.11 40.02 40.09 40.14 39.93 40.05 39.88 40.04 39.95 ~X R 0.38 0.19 0.36 0.22 0.26 0.24 0.12 0.3 0.14 0.17 40.08 40.01 40.05 40.02 40.07 39.94 40 39.93 40.05 39.98 R = 2.38/10 = 0.238

~ CL= X = 40.013 ~ UCL= X+m3 A2R= 40.013 +0.691×0.238 =40.18 ~ LCL= X- m3 A2R= 40.013 -0.691×0.238 =39.85 CLR=R= 0.238

UCLR= D4R=2.115×0.238 = 0.5 LCLR= D3R=0×0.238 = 0 轮廓线： UCL = 40.18

CL = 40.013 LCL= 39.85

UCLR= 0.5 CLR= 0.238 LCLR= 0 35. 解：p =

?Pn = 156/13670 = 1.14% ?nn≤ni≤2n 2 n = 683. 5 pn= 7.8≥3 (近似于正态分布)

CL= p = 1.14% UCL= p +3

p(1?p)n =0.0263 LCL= p -3

p(1?p)n= - 0.0008 (取LCL=0 ) 轮廓线： UCL = 0.0263

CL = 0.0114

LCL= 0

36. 解：

p =

?Pn?n = 66/1650 = 4 % n = 82. 5 pn= 3.3≥3 (近似于正态分布) CL= p = 4 % UCL= p +3

p(1?p)n =0.105 LCL= p -3

p(1?p)n= - 0.025 (取LCL=0 )

轮廓线： UCL = 0.105

CL = 0.04

LCL= 0

37. 解：p =

?Pn?n = 60/3000 = 0.02 pn = 6≥3 (近似于正态分布) CL= pn =0.02×300 = 6 UCL= pn +3pn(1?p) =13.27

n2≤ni≤2n LCL= pn -3pn(1?p) = - 1.27 (取LCL=0 ) 轮廓线： UCL = 13.27

CL = 6

LCL= 0

38. 解：p =

?Pn = 40/1000 = 0.04 ?n pn = 4≥3 (近似于正态分布) CL= pn =0.04×100 = 4 UCL= pn +3pn(1?p) =9.98

LCL= pn -3pn(1?p) = - 1.88 (取LCL=0 )

轮廓线： UCL = 9.98

CL = 4

LCL= 0

139. 解：C =?C = 60/10 = 6 ≥3 (近似于正态分布)

K CL= C = 6

UCL= C +3c = 13.35

LCL= C -3c = - 1.35 (取LCL=0 )

轮廓线： UCL = 13.35

CL = 6

LCL= 0

140. 解：C =?C = 48/10 = 4.8 ≥3 (近似于正态分布)

K CL= C = 4.8

UCL= C +3c = 11.37

LCL= C -3c = - 1.77 (取LCL=0 ) 轮廓线： UCL = 11.37

CL = 4.8

LCL= 0

c?41. 解：u= = 52/17.5 = 2.97≈3 (近似于正态分布) n? CL= u = 2.97 n =

?n = 17.5 =1.75 （

k10n≤ni≤2n ） 2 ∴ U CL= u+3

u= 6.8782 nu = - 0.9382 (取LCL=0 ) n LCL= u-3

轮廓线： UCL = 6.8782

CL = 2.97

LCL= 0

42. 解：计算出Xcs 、Rs见上表

1s?X=0.06 /10= 0.006 Xc=n Rs=

?Rs = 0.474/9 =0.0527

N?1 CL= Xcs= 0.006

UCL= Xcs + E2Rs = 0.006 +2.66×0.0527= 0.1461 LCL= Xcs- E2Rs = 0.006 -2.66×0.0527= -0.1341