What are the mean and standard deviation of a binomial probability distribution with n=20 and...

Mean 20
Standard deviation ${\displaystyle \frac{3}{\sqrt{5}}}$
Explanation:
${\displaystyle \mu =np}$
=(20)*(0.9)
=18
${\displaystyle \sigma =\sqrt{np(1-p)}}$
${\displaystyle =\sqrt{20\cdot \left(0.9\right)\cdot \left(0.1\right)}}$
${\displaystyle =\frac{3}{\sqrt{5}}}$
If the random variable X follows the binomial distribution with parameters ${\displaystyle n\in \mathbb{N}}$ and ${\displaystyle p\in [0,1]}$, tthe probability of getting exactly k successes in n trials, is given by
${\displaystyle Pr(X=k)=n{C}_k\cdot p^k\cdot {(1-p)}^{n-k}}$
The mean is calculated from E(X).
The standard deviation comes from the square root of the variance,
${\displaystyle Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2}$
Where
${\displaystyle E\left(X\right)=\sum _{k=0}^{n}k\cdot Pr(X=k)}$
${\displaystyle E\left(X^2\right)=\sum _{k=0}^n{k}^2\cdot Pr(X=k)}$

The mean and standard deviation of a binomial probability distribution with
n=200....(1)
p=0.3....(2)
can be determined as follows.
We know that the mean of a binomial probability distribution is given as:
${\displaystyle \mu =np}$...(3)
And the standard deviation of a binomial probability distribution is given as:
${\displaystyle \sigma =\sqrt{np(1-p)}}$...(4)
Substituting (1) and (2) in (3), we get:
${\displaystyle \mu =200\left(0.3\right)}$
${\displaystyle \mu =60}$
Substituting (1) and (2) in (4), we get:
${\displaystyle \sigma =\sqrt{200\left(0.3\right)(1-0.3)}}$
${\displaystyle =\sqrt{200\left(0.3\right)\left(0.7\right)}}$
${\displaystyle =\sqrt{42}}$
${\displaystyle \sigma =6.48}$