a. Expected value is defined by
[tex]E[X]=\displaystyle\sum_xx\,p(x)[/tex]
so we get
[tex]E[X]=1\cdot0.05+2\cdot0.10+4\cdot0.30+8\cdot0.45+16\cdot0.10[/tex]
[tex]\boxed{E[X]=6.65}[/tex]
b. Variance is defined by
[tex]V[X]=E[(X-E[X])^2][/tex]
so with the expectation found above, we have
[tex]V[X]=E[(X-6.65)^2][/tex]
[tex]V[X]=\displaystyle\sum_x(x-6.65)^2\,p(x)[/tex]
(by definition of expectation)
[tex]V[X]=(1-6.65)^2\cdot0.05+(2-6.65)^2\cdot0.10+(4-6.65)^2\cdot0.30+(8-6.65)^2\cdot0.45+(16-6.65)^2\cdot0.10[/tex]
[tex]\boxed{V[X]=15.4275}[/tex]
c. Standard deviation is the square root of variance:
[tex]\boxed{\sqrt{V[X]}\approx3.928}[/tex]
d. I assume "shortcut formula" refers to
[tex]V[X]=E[X^2]-E[X]^2[/tex]
which is easily derived from the definition of variance. We have (by def. of expectation)
[tex]E[X^2]=\displaystyle\sum_xx^2\,p(x)[/tex]
[tex]E[X^2]=1^2\cdot0.05+2^2\cdot0.10+4^2\cdot0.30+8^2\cdot0.45+16^2\cdot0.10[/tex]
[tex]E[X^2]=59.65[/tex]
and so the variance is again
[tex]V[X]=59.65-6.65^2[/tex]
[tex]\boxed{V[X]=15.4275}[/tex]
as expected.
