Is it possible to find the x value for a given y value of a fitdist() distribution?

Hi,

I'm using a probplot to display 500ish data points. I have this setup in a AppDesigner app. A distribution line can be added using the fitdist function as show below

DataType = app.DataTypeDropDown.Value;
    dataFit = app.SelectFitDropDown.Value;
DistributionType = app.DistributionDropDown.Value;
    device = app.DeviceDropDown.Value;

LegendName1 = ' Distribution Fit';
LegendName2 = ' ';
legendname = append(device,LegendName2,dataFit,LegendName1);

z = app.distData(:,device);
array = table2array(z);
array = str2double(array);

pd = fitdist(array(:,1),dataFit);
pp = probplot(app.UIAxes,pd);
set(pp,'color',app.colourMatrix(app.colourCounter,:),'DisplayName', legendname,'LineWidth', 1.5);
pp.ShowPoles = 'off';

When the fitted distribution is fitted, I would like to add a straight vertical line through the fitted distribution for a given Y value.

Is it possible to do this using the fitdist distribution or similar?

[dpb]

What is "Y", here?

You can find P(X) or the inverse P() for the fitted distribution of choice through the associated helper functions but it isn't clear which it is you're after here...

[Andrew Feenan]

y would be 0.99999. I just need one X value from the fitted distribution

[dpb]

I presume that 5-9's is then a tail (upper) probability value?

In that case, then the icdf function for the named distribution is what you're looking for.

[Andrew Feenan]

Thank you, that has helped me a lot.

Answer 1

[Andrew Feenan]

@dpb provied the solution to my question in the comments above.

I presume that 5-9's is then a tail (upper) probability value?

In that case, then the icdf function for the named distribution is what you're looking for.