How is this Variance found in this old question?

Question

On this question: Probability: Normal Distribution they find these values: $\hat\mu = .05(150) = 7.5\space,\hat\sigma = \sqrt{150(.05)(.95)} = 2.67$

I see how they got $\mu$, but how did they get $\sigma$ ? Are the "hats" a referring to a special kind of these variables? This is the only way I know to compute $Var(X)$: $E[X^2]-E[X]^2$, and that gives

$\sqrt{0.5*150^2-(0.5*150)^2} = 32.69$

I know I'm wrong, as I have found the same solution elsewhere, but what am I missing?

Answer 1

This question was sitting among a host of normal-distribution questions, so I automatically tried to use the normal distribution Variance equation. The Variance equation for a binomial random variable is

$Var(X) = n*p*(1-p)$

which is the correct equation to use. Sorry about that, I'm still new to this area of mathematics and didn't even realize there were multiple versions.