I have peak value of normal distribution $0.581$ I know mean which is $0.01806$. I want to find variance now. But I know value at a certain point for continuous distribution is zero. How will I do Integration?

$\dfrac{1}{\sqrt{2 \pi} \sigma }e^{\frac{\sum_i(x_i-\mu)^2}{\sigma^2}}= \dfrac{1}{\sqrt{2 \pi} \sigma }e^{\frac{\sum_i(x_i-0.01806)}{\sigma^2}^2 } $

I am not sure how to proceed. I need hint or somthing


I assume that you mean that the maximum value of the pdf of the normal distribution is $0.581$. For the normal distribution, the maximum is attained at the mean. So we have $f(x=\mu;\mu, \sigma)=0.581=f(0.01806;0.01806, \sigma)$, where $f(x;\mu, \sigma)$ denotes the pdf of the normal distribution with mean $\mu$ and standard deviation $\sigma$. All you have to do now is to plug those values in the formula and solve for $\sigma$. As a reminder, the pdf of the normal distribution is $$ f(x;\mu,\sigma)=\frac{1}{\sigma\sqrt{2\pi}}\,\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}\right) $$

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    $\begingroup$ ohh I see now so we will be left with $\dfrac{1}{\sqrt{2 \pi} \sigma }$ right ? $\endgroup$ – Daman Jul 30 at 10:07
  • $\begingroup$ Yes, that's right! $\endgroup$ – COOLSerdash Jul 30 at 10:07

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