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How to solve ncdf(X, 0.76, 83, 1E99) = 0.15 (RESOLVED)
04-27-2015, 01:41 PM (This post was last modified: 04-28-2015 04:08 PM by StephanP.)
Post: #1
How to solve ncdf(X, 0.76, 83, 1E99) = 0.15 (RESOLVED)
While evaluating the HP Prime, I am trying to sort out how to do the things I usually do on my TI-84.

I can't seem to find how to solve the following equation:

ncdf(X, 0.76, 83, 1E99) = 0.15

normald_CDF( u0=X, sd=-.76, left bound=83, right bound = 1E99)
area (probability) = 0.15

On TI-84 I would do the following:
Y1 = normalcdf(83, 1E99, X, 0.76)
Y2 = 0.15
Adjust window settings, Graph and Calculate intersect

   

How would I do this on the HP Prime?
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04-27-2015, 02:33 PM (This post was last modified: 04-27-2015 02:38 PM by Tim Wessman.)
Post: #2
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
Simpler actually (although you could do it the same way if you really wanted). NOTE: I can add some pictures later if you want.

Open solve app. Remember that most applications have a symbolic, plot and numerical representation. Pressing the Symb, Plot or Numeric buttons will switch between them.

In Symb view, type NORMALD_CDF(X,.76,83,1E99)=.15 (the "maximum" real value is actually 1E499, so you could replace "1E99" by "MAXREAL" if desired)

The press Num to go to the solver. Press SOLVE on the screen button to solve numerically. Plot will open a plot and you can view the crossing point if desired with a bit of scrolling/adjustment/scaling.

Now if you want to do the exact same way with the function application, you can do so. Just define F1(X) as NORMALD_CDF(X,.76,83,MAXREAL) and F2(X)=.15. Plot to see the plot, Menu button on the screen -> Fnc -> Intersection to jump to your intersection (no left/right needed). Did you want to look at a table instead and "manually" figure out the X value? Well, just press num. You can see the value of F1(X). Type a number to jump to that location. As you approach your value, you will see the values approaching 1 the closer you get.


You also have the option of defining your solve equation, E1 as NORMALD_CDF(X,S,L,U)=P. Then you will get 4 variables to fill in with desired values and solve for the last one. Very flexible. :-)

TW

Although I work for the HP calculator group, the views and opinions I post here are my own.
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04-27-2015, 02:43 PM
Post: #3
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
Yeah, it DOES work.
I must have made some typo somewhere.
And how fast is that intersection calculated. ARM Powered!

I'll delve into the SOLVE App right away. (Something new to experience)
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04-27-2015, 02:47 PM
Post: #4
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
Oh, that SOLVE App is a nice one too.
So fast.

Issue resolved (in two ways)
Thanks, again.
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04-27-2015, 05:02 PM
Post: #5
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
Thanks Stephen for the new function.

It seems you can also use solve in CAS for this:

solve((normald_cdf(83,0.76,x)) = 0.15,x)
{82.212310624}
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04-27-2015, 07:23 PM
Post: #6
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
(04-27-2015 05:02 PM)CR Haeger Wrote:  Thanks Stephen for the new function.

It seems you can also use solve in CAS for this:

solve((normald_cdf(83,0.76,x)) = 0.15,x)
{82.212310624}

Hmmm... does the Prime really have no quantile functions that give a direct result?
Here's how you do it on a 34s:

83 STO J   0,76 STO K
0,15 Norml–1 => 82,21231062398472

Dieter
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04-27-2015, 08:18 PM
Post: #7
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
(04-27-2015 07:23 PM)Dieter Wrote:  Hmmm... does the Prime really have no quantile functions that give a direct result?
Here's how you do it on a 34s:

83 STO J   0,76 STO K
0,15 Norml–1 => 82,21231062398472

Dieter

The Prime can solve directly for x with: NORMALD_ICDF(83,0.76,0.15)

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.
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04-28-2015, 06:28 AM
Post: #8
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
(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
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04-28-2015, 06:32 AM
Post: #9
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
(04-27-2015 02:33 PM)Tim Wessman Wrote:  In Symb view, type NORMALD_CDF(X,.76,83,1E99)=.15 (the "maximum" real value is actually 1E499, so you could replace "1E99" by "MAXREAL" if desired)

1E+99? Or even MAXREAL=9,9999...E+499?

The Normal CDF drops below 1E–500 before z reaches 50 (!). For most applications setting the limit to z=8 is more than sufficient – here the upper CDF is less than 1E–15.

Dieter
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04-28-2015, 11:32 AM (This post was last modified: 04-28-2015 11:34 AM by CR Haeger.)
Post: #10
RE: How to solve ncdf(X, 0.76, 83, 1E99) = 0.15
(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 Smile.
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