高中数学：2.2《总体分布估计》苏教必修3

总体分布.抽样序号n=25n=250n=2500mm/nmm/nmm/n110.04120.0481570.0628240.16140.0561520.0608300.00170.0681570.0628400.00110.0441360.0544510.04220.0881520.0608610.0490.0361350.0540720.08150.0601430.0572800.00140.0561600.0640910.04210.0841490.05961010.0480.0321530.0612.试验结果概率次品0.06正品0.94.试验结果频数频率国徽向上361240.5011国徽向下359640.4989样本（容量72088）频率分布试验结果概率国徽向上0.5国徽向下0.5总体分布.次品正品试验结果0.93880.0612频率.从规定尺寸为25.40mm的一堆产品中任取100件，测得它们的实际尺寸如下：25.3925.3625.3425.4225.4525.3825.3925.4225.4725.3525.4125.4325.4425.4825.4525.4325.4625.4025.5125.4525.4025.3925.4125.3625.3825.3125.5625.4325.4025.3825.3725.4425.3325.4625.4025.4925.3425.4225.5025.3725.3525.3225.4525.4025.2725.4325.5425.3925.4525.4325.4025.4325.4425.4125.5325.3725.3825.2425.4425.4025.3625.4225.3925.4625.3825.3525.3125.3425.4025.3625.4125.3225.3825.4225.4025.3325.3725.4125.4925.3525.4725.3425.3025.3925.3625.4625.2925.4025.3725.3325.4025.3525.4125.3725.4725.3925.4225.4725.3825.39请问：（1）这个问题的“总体”是什么？（2）如何处理这些数据？即如何估计这个问题的“总体分布”？.分组个数累计频数频率[25.235,25.265)一10.01[25.265,25.295)┰20.03[25.295,25.325)正50.05[25.325,25.355)正正┰120.12[25.355,25.385)正正正下180.18[25.385,25.415)正正正正正250.25[25.415,25.445)正正正一160.16[25.445,25.475)正正下130.13[25.475,25.505)40.04[25.505,25.535)丅20.02[25.535,25.565)丅20.02合计

1001.00第六步，列出频率分布表。第六步，列出频率分布表。分组个数累计频数频率[25.235,25.265)一10.01[25.265,25.295)┰20.03[25.295,25.325)正50.05[25.325,25.355)正正┰120.12[25.355,25.385)正正正下180.18[25.385,25.415)正正正正正250.25[25.415,25.445)正正正一160.16[25.445,25.475)正正下130.13[25.475,25.505)40.04[25.505,25.535)丅20.02[25.535,25.565)丅20.02合计

1001.00正.25.23525.29525.35525.41525.47525.535产品尺寸(mm）频率分布直方图.25.23525.29525.35525.41525.47525.535产品尺寸(mm）总体密度曲线.为了了解一大片经济树林的生长情况，随机测量其中的100株的底部周长，得到如下数据表（长度单位：cm）135981021109912111096100103125971171131109210210910411210912487131971021231041041281051231111031059211410810410212912697100115111106117104109111891101218012012110410811812999909912112310711191100991011169710210810195107101102108117991181061199712610811998