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如何找到2个骰子的以下属性? (里面的细节)
“a)0.351087”“b)7.2”“c)0.056627”“P [sum is 8] = 5/36”“因为有5种可能的组合投掷8:”“(2,6),(3,5) ),(4,4),(5,3)和(6,2)。“ “a)这相当于我们连续7次获得”和“不等于8的总和的几率,这些是”(1 - 5/36)^ 7 =(31/36)^ 7 = 0.351087“b )36/5 = 7.2“”c)“P [”x = 8 | x> = 2“] =(P [”x = 8,x> = 2“])/(P [”x> = 2“ ])=(P [“x = 8”])/(P [“x> = 2”])P [“x = 8”] = 0.351087 *(5/36)= 0.048762 P [“x> = 2 “] = P [”第一次总和不是8“] = 31/36 => P [”x = 8 | x> = 2“] = 0.048762 * 36/31 = 0.056627
我如何计算给定的事件? (里面的细节,对我来说有点复杂)
“见解释”“y是标准正常(平均值为0,标准差为1)”“所以我们使用这个事实。” “1)”= P [ - 1 <=(xz)/ 2 <= 2]“我们现在在表格中查找”z = 2和z = -1的z值的z值。我们得到“0.9772 “和”0.1587。 => P = 0.9772-0.1587 = 0.8185“2)”var = E [x ^ 2] - (E [x])^ 2 => E [x ^ 2] = var +(E [x])^ 2“这里我们有var = 1和mean = E [Y] = 0。“ => E [Y ^ 2] = 1 + 0 ^ 2 = 1“3)”P [Y <= a | B] =(P [Y <= a“AND”B])/(P [B])P [B] = 0.8413-0.1587 = 0.6826“(z值表)”P [Y <= a“AND”B ] = 0,“if”a <-1 P [Y <= a“AND”B] = P [-1 <= Y <= a]“,如果”-1 <= a <= 1 P [Y < =“AND”B] = P [B]“,如果”a> 1 => P [Y <= a | B] = 0,“if”a < - 1 => P [Y <= a | B] =(T(a)-0.1587)/0.6826,“if”-1 <= a &