For X~N(μ,σ^2), find k such that:
Yes...I know ilovemadonna2009's answer is not accurate since the answer is just obtained by observation.
- doraemonpaulLv 71 十年前最愛解答
設h = k/σ，則此方程會變成P(-2h < Z < h) = 0.9（亦由此證出P(-2k/σ < Z < k/σ) = 0.9的解是形如k = hσ，其中h是常數的形式）。
2010-03-05 16:09:43 補充：
用normal distribution table的解法：
找h1和h2使P(-2h1 < Z < h1) < 0.9 < P(-2h2 < Z < h2)，然後進行interpolation。但這解法的準確度較差。
2010-03-05 16:09:57 補充：
所以，較好的解法當然是用Newton's method（fix-point iteration似乎不太可行）解∫(-2h to h) 1/√2π e^(-x²/2) dx = 0.9，可用fx-3650P幫手，這至少可以解出一個準確至5位小數的近似解。（∫(-2h to h) 1/√2π e^(-x²/2) dx = 0.9亦是方程一條！）
唯一留意的是，d/dh(∫(-2h to h) 1/√2π e^(-x²/2) dx - 0.9)需打醒十二分精神，千萬不要d錯！
2010-03-09 07:01:32 補充：
2010-03-11 08:36:38 補充：
my wisdom of maths
- 1 十年前
- 自由自在Lv 71 十年前
若以Normal Distribution Table可得k大約於1.307及1.308之間.
匿名提供之Excel Goal Seek 也只是約數1.3074.其實可以Excel函數Norminv及Normdist
2010-03-06 14:51:41 補充：
2010-03-13 13:53:23 補充：
Excel Goal Seek is a good method to solve the problem since in all practical cases for statistics, we do not need to find a very accurate answer. So in this sense, Excel Goal Seek is not at all a junk solution. It is simple. I am here to share my another thought and a precise figure I can get.
- 匿名Lv 61 十年前
The method by ilovemadonna2009 is right, but the answer is wrong
Actually,the required range is not symetrical. How could we estimate the answer by symetry
With the goal seek function in excel, the numerical answer is 1.3074
2010-03-10 08:38:30 補充：
nelsonywm2000, your method (i.e. Newton's Method) is also one of the numerical method. It seems it not much better than using Excel to solve it.
Do you have any analytical method to find the exact answer? e.g. using integration?
2010-03-12 08:35:54 補充：
doraemonpaul, it seems that the difference is only the degree of accuracy. All we used / proposed / suggested are numerical methods. There are a thousand of different numerical methods available, depending on the accuracy and efficiency.
2010-03-12 08:36:11 補充：
unless there is an analytical solution
2010-03-12 08:39:07 補充：
Actually, you could use Simpson's Rule to evaluate the definite intregal, rather than use Newton's Method. it is much more direct and appropriate.