chapter 1

Chapter 16: Quality Control MethodsLTPDAQLnp= p.07.0234. = = 3.5 ≈ 3. 55 , which appears in the1pp2.613= 130.65 ≈131.025⎛131⎞xp = .02) = 1−∑⎜⎟ .02 .98x=0⎝x ⎠5⎛131⎞x 131−p = .07) = ∑⎜⎟ .07 .93x=0⎝x ⎠1n =.P ( X > 5 when ( ) ( )P ( X ≤ 5 when ( ) ( )12column in the c = 5 row. Then131−xx= .0487 ≈ .05= .0974 ≈ .1035. P(accepting the lot) = P(X 1 = 0 or 1) + P(X 1 = 2, X 2 = 0, 1, 2, or 3) + P(X 1 = 3, X 2 = 0, 1, or 2)= P(X 1 = 0 or 1) + P(X 1 = 2)P(X 2 = 0, 1, 2, or 3) + P(X 1 = 3)P( X 2 = 0, 1, or 2).p = .01: = . 9106 + (.0756)(.9984) + (.0122)(.9862) = . 9981p = .05: = . 2794 + (.2611)(.7604) + (.2199)(.5405) = . 5968p = .10: = . 0338 + (.0779)(.2503) + (.1386)(.1117) = . 068836. P(accepting the lot) = P(X 1 = 0 or 1) + P(X 1 = 2, X 2 = 0 or 1) + P(X 1 = 3, X 2 = 0) [since c 2 = r 1– 1 = 3] = P(X 1 = 0 or 1) + P(X 1 = 2)P( X 2 = 0 or 1) + P(X 1 = 3)P(X 2 = 0)1⎛50⎞⎛501⎞⎛100x⎞= ∑⎜⎟p∑x=0⎝x ⎠⎝ 2 ⎠x=0⎝x ⎠⎛50⎞⎛100⎞= ⎜ ⎟p3 ( 1−p) 47⋅⎜⎟p0 ( 1−p) 100.⎝ 3 ⎠ ⎝ 0 ⎠= . 7358 + .1858 .4033 + .0607 .1326 = .= . 2794 + .2611 .0371 + .2199 .0059 = .= . 0338 + .0779 .0003 + .1386 .0000 = .50−x2 48x( 1−p) + ⎜ ⎟p( 1−p) ⋅ ⎜ ⎟ p ( 1−p)p = .02: ( )( ) ( )( ) 8188p = .05: ( )( ) ( )( ) 2904p = .10: ( )( ) ( )( ) 0038100−x478