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what is the best way to find a function that looks like a normal distribution, when the curve intersects the y-axis above 0 (say 0.3) and the area must be 1 (100%)?

https://i.stack.imgur.com/5cJSy.jpg

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The PDF of the normal distribution is $\frac 1 {\sqrt{2 \pi \sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$. If $\mu=0$, then the PDF at zero is $\frac 1 {\sqrt{2 \pi \sigma^2}}$. So if you want the y-intercept to be $y_0$, you just need to solve $y_0=\frac 1 {\sqrt{2 \pi \sigma^2}}$ for $\sigma$. That gives $\frac 1{y_0^2}=2\pi\sigma^2\rightarrow\sigma^2=\frac1{{2\pi y_0^2}}\rightarrow \sigma = \frac{1}{y_0 \sqrt{2\pi}}$ for $y_0>0$.

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  • $\begingroup$ Thank you very much. I'm making a thinking error. I need to express that the probability of spending 0 (x) is 0.3 (y), and the probability of spending less than 0 (no negative numbers on the x-axis) is 0 (y). from there I will find the CDF. $\endgroup$ – Phi Al May 9 at 9:58

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