我有一个Zigma值,但我如何使用.Normaldistribution找到Z值 [英] I Have A Zigma Value But How Can I Use.Normaldistribution To Find The Z Value

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问题描述

private void findSumvarience()
        {

            for (int i = 0; i < CriticalPath.Count; i++)
            {
                for (int j = 0; j < Proj_name_Value.Count; j++)
                {
                    if (CriticalPath[i] == Proj_name_Value[j])
                    {

                        Varience1 = variance[j] + Varience1;

                    }
                }
            }
            Varience1 = Convert.ToDecimal(Math.Sqrt(Convert.ToDouble(Varience1)));
            Varience1 = Math.Round(Varience1, 2, MidpointRounding.AwayFromZero);
            //Weak_Guest.Text = Convert.ToString(Varience1);
        }







string Check;
           Check = Weak_Guest.Text;
           //MessageBox.Show(Convert.ToString(Check.Length));
           if (Check.Length == 0 || Weak_Guest.Text == " ")
           {
               MessageBox.Show("PLease Insert Max Weak TO Culculate");
           }
           else
           {
               WeakToCal = Convert.ToInt32(Check);
               if (WeakToCal < Time)
               {
                   MessageBox.Show("Project can't Finish in Time Becuare Max Weak Is Less Than Time");
               }
               else
               {
                   Result = WeakToCal - Time;

                   Result = Result / Varience1;
                   Result = Math.Round(Result, 2, MidpointRounding.AwayFromZero);
                   //Result = Chart.da
                   MessageBox.Show(Convert.ToString(Result));
               }

           }





i Gor a zigma并获得用户想要计算的弱点textbok名称weakguest的可能性。找到可能性我需要通过比较任何人知道如何做到的Zigma值来看看normaldistribution的可能性是什么?请



PS。如果我的英语很糟糕我很抱歉。



i Gor a zigma and get the weak that user want to calculate the posibility by textbok name weakguest . to find posibility i need to see what is the posibility in normaldistribution by compare the Zigma value any one know how to do that ? Please

PS. IF my English Suck i am sorry.

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