《质量管理学》课程综合练习题(9)

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

轮廓线： UCL = 0.1461

CL = 0.006

LCL= -0.1341

43. 解：(1)图 6 12.6 5 10.4 8 18.5 1 3 4 8.1 7 16.3 6 12.3 3 6.2 3 6.6 7 16.8 (2)从散点图可以发现每人月平均销售额和利润率之间存在着线性关系，而且相关程度较高，可以配合直线回归方程。

(3)计算相关系数，先列出计算过程所需要的有关数字(表见下页)。r=0.9874，表明每人平均月销售额和利润率之间存在着显著的正相关关系。

(4) n x y xx xy 1 6 12．6 36 75 2 5 10．4 25 52 3 8 18．5 64 148 4 1 3．0 1 3 5 4 8．1 16 32．4 6 7 16．3 49 114．1 7 6 12，3 36 73．8 8 3 6，2 9 18．6 9 3 6．6 9 19．8 10 7 16．8 49 117．6 合计 50 110．8 294 654．9 b =

?XY?X?Y = ?X?X?X26549?50?110.8 = 2.293

2940?2500a = Y - bX= 11.08 - 2.293×5 = - 0.385 ∴回归方程为：y = - 0.385+2.293x (5) Yc = - 0.385+2.293×2 =4.2