Q.The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of $n$ and $p$ are :

Four named discrete distributions, in order of increasing generality, each with its own closed-form mean and variance:

• One-point distribution ($f(x_0)=1$): mean $=x_0$, variance $=0$ — no spread is possible when all mass sits at one value (matches Corollary 4 of the expectation properties, $V(b)=0$).
• Two-point distribution (values $x_1,x_2$ with probabilities $p,q=1-p$): mean $=px_1+qx_2$, variance $=pq(x_2-x_1)^2$; in the symmetric case $p=q=\tfrac12$ these become $\dfrac{x_1+x_2}{2}$ and $\dfrac{(x_2-x_1)^2}{4}$.
• Bernoulli distribution $X\sim\text{Ber}(p)$ (the two-point distribution specialised to $x_1=1,x_2=0$): mean $=p$, variance $=pq$; symmetric case ($p=\tfrac12$) gives mean $\tfrac12$, variance $\tfrac14$.
• Binomial distribution $X\sim B(n,p)$ (the sum of $n$ independent Bernoulli$(p)$ trials): mean $=np$, variance $=np(1-p)=npq$; symmetric case ($p=\tfrac12$) gives mean $\dfrac n2$, variance $\dfrac n4$. …