How to solve ncdf(X, 0.76, 83, 1E99) = 0.15 (RESOLVED)

[CR Haeger]

(04-28-2015 06:28 AM)Dieter Wrote:  

(04-27-2015 08:18 PM)CR Haeger Wrote:  The Prime can solve directly for x with: NORMALD_ICDF(83,0.76,0.15)

Great. But why then would one want to determine a quantile by solving the CDF? Sounds a bit like calculating sqrt(2) by solving x²–2 = 0. #-) But maybe I didn't get the essential point here...

(04-27-2015 08:18 PM)CR Haeger Wrote:  In this example, the WP34S used ~16-19 keystrokes depending on PROB menu navigation. The Prime takes ~14 or more depending on having to go to the on-board HELP to guide the syntax to use.

On the 34s I would actually do it this way:

,15  g Φ^–1  ,76 x 83 +   => 82,21231...

That's 12 keystrokes.
And no need to consult the manual to find out where to store µ and σ. ;-)

Dieter


Dieter

I showed the CAS solve() example only to illustrate that you could use it to solve for any of the four CDF variables - I just chose x in the example. I'm not suggesting its the "best" method.

I agree that the WP34S is in many ways more straightforward/efficient. If you were lucky enough to have J, K stored already and had been using Norml^–1 recently, then maybe you're down to only five keystrokes .