Given, \[X\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\]

P(X = r) \[={{~}^{n}}{{C}_{r}}~{{p}^{r}}~{{q}^{n-r}}\]

Where \[n\text{ }=\text{ }3,\text{ }p\text{ }=\text{ 1/2},\text{ }q\text{ }=\text{ 1/2}\]and \[r\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\]

\[P\left( X\text{ }=\text{ }0 \right)\text{ }=\text{ 1/2}\times \text{1/2}\times \text{1/2 }=\text{ }1/8\]

\[P\left( X\text{ }=\text{ }1 \right)\text{ }=\text{ }3\text{ }\times \text{1/2}\times \text{ 1/2 }\times \text{1/2 }=\text{ }3/8\]

\[P\left( X\text{ }=\text{ }2 \right)\text{ }=\text{ }3\text{ }\times \text{ 1/2}\times 1/2\times 1/2\text{ }=\text{ }3/8\]

\[P\left( X\text{ }=\text{ }3 \right)\text{ }=\text{ 1/2}\times \text{1/2}\times \text{1/2 }=\text{ }3/8\]

The probability distribution table is:

Now,

\[E\left( X \right)\text{ }=\text{ }0\text{ }+\text{ }1\text{ }\times 3/8\text{ }+\text{ }2\text{ }\times 3/8\text{ }+\text{ }3\text{ }\times \text{ }1/8\text{ }=\text{ }3/8\text{ }+\text{ }6/8\text{ }+\text{ }3/8\text{ }=\text{ }12/8\text{ }=\text{ }3/2\]

\[E({{X}^{2}})\text{ }=\text{ }0\text{ }+\text{ }1\text{ }\times 3/8\text{ }+\text{ }4\text{ }\times 3/8\text{ }+\text{ }9\text{ }\times 1/8\text{ }=\text{ }3/8\text{ }+\text{ }12/8\text{ }+\text{ }9/8\text{ }=\text{ }24/8\text{ }=\text{ }3\]

W.k.t, \[Var\left( X \right)\text{ }=\text{ }E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}~=\text{ }3\text{ }\text{ }{{\left( 3/2 \right)}^{2}}~=\text{ }3\text{ }\text{ }{{\left( 3/2 \right)}^{2}}~=\text{ }3\text{ }\text{ }9/4\text{ }=\text{ 3/4}\]

Thus, the standard deviation = Var(X) = \[\sqrt{3/4}\text{ }=\text{ }\sqrt{3}/2\]