# Variance of Normal distribution given all values

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 $$f(x;\mu,\sigma)=\frac{1}{\sigma\sqrt{2\pi}}\,\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}\right)$$
• ohh I see now so we will be left with $\dfrac{1}{\sqrt{2 \pi} \sigma }$ right ? – Daman Jul 30 at 10:07