How do I calculate standard deviation and $\bar{x}$ if radius of circle is given?

Question

It is given radius of circle $r = 5cm$ and standard deviation $\sigma = 1mm$. How do I calculate $\bar{x}$ and standard deviation of calculated circle area?

I know these formulas but I dont know how to use them:

$\bar{x}=\frac{1}{N}\sum_{N}^{i=1}x_i$

$\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{N-1}}$

What is $N$ and $x_i$ in my case?

Answer 1

The area is $A=\pi r^2$

so $\mathbb E[A] = \pi\mathbb E[r^2] =\pi((E[r])^2+\sigma_r^2) = 25.01 \pi \approx 78.5712 \text{ cm}^2$

The standard deviation of $A$ depends on the distribution of $r$. You could calculate it as $\sqrt{\mathbb E[A^2]-(E[A])^2} = \pi \sqrt{\mathbb E[r^4]-(E[r^2])^2}$. For example

• $r$ might take the values $4.9$ and $5.1\text{ cm}$ with equal probability, in which case the standard deviation of $A$ would be $\pi \approx 3.1416 \text{ cm}^2$. I suspect that this may be the minimum possible value
• $r$ might be distributed with zero skewness and excess kurtosis (like a normal distribution, though here $r$ cannot be negative) in which case the standard deviation of $A$ would be marginally greater at about $3.1419 \text{ cm}^2$