A random variable following a binomial distribution over \(n\) trials with success probability \(p\) has PMF
\(f_X(x)=\dbinom nxp^x(1-p)^{n-x}\)
Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.
\(\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1\)
The mean is given by the expected value of the distribution,
\(\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}\)
\(\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\)
\(\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\)
\(\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}\)
\(\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}\)
\(\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}\)
\(\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}\)
The remaining sum has a summand which is the PMF of yet another binomial distribution with \(n-1\) trials and the same success probability, so the sum is 1 and you're left with
\(\mathbb E(x)=np=126\times0.27=34.02\)
You can similarly derive the variance by computing \(\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2\), but I'll leave that as an exercise for you. You would find that \(\mathbb V(X)=np(1-p)\), so the variance here would be
\(\mathbb V(X)=125\times0.27\times0.73=24.8346\)
The standard deviation is just the square root of the variance, which is
\(\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834\)
