广西大学：《概率论与数理统计》课程电子教案（PPT课件）第七章 参数估计 7.6 补充例题

7.6（0-1）分布参数的区向估针补充1设从一大批产品的120个样品中，得次品9个，求这批产品的次品率p的置信度为0.90的置信区间9解n =120, 0.075, 1-α = 0.90x=120Zα/2 = Z0.05 = 1.645,则a = n + za/2 = 122.71,b = -(2nX + za/2) = -(2nx + za/2) = -20.71,K

得次品 解 n = 120, 0.075, 120 9 x = = 1− = 0.90, 1.645, z / 2 = z0.05 = 122.71, 2 则a = n + z / 2 = (2 ) 2  / 2 b = − nX + z (2 ) 2  / 2 = − nx + z = −20.71, 置信区间. 求这批产品的次品率 p 的置信度为0.90的 设从一大批产品的120个样品中, 9个, 补充1

7.6（0一1）分布参数的区向估计c = nX2 = nx2 = 0.675,- b-~/b?- 4ac= 0.044,于是 p =2a-b+/b2-4ac= 0.125,P2 =2ap的置信度为0.90的置信区间为（0.044,0.125)K

2 2 c = nX = nx = 0.675, a b b ac p 2 4 2 1 − − − 于是 = a b b ac p 2 4 2 2 − + − = = 0.044, = 0.125, p 的置信度为0.90的置信区间为 (0.044, 0.125)