Probability using standard deviation !Probability using standard deviation The current through a 1 k resistor is a Gaussian random variable with mean 10 and standard deviation 6 (in milliamperes). Calculate the probability that the power dissipated by the resistor does not exceed 100 milliwatts.

First, the power equation can be written as $$ P = I^2 R $$ where you have $R = 1000 \Omega$ and $I$ is given by $\mu = 10mA$, $\sigma = 6mA$. The power exceeds $100 mW$ when $ P = I^2 R \geq 0.1 W$ or in other words, when $$ I \geq \sqrt{\frac{1000}{0.1}} = \sqrt{10000} = 100 A $$ So the question is, what is the probability that $I \geq 100 A$.

This makes me think that maybe your units are off, because that probability is effectively zero...