The PMF of [tex]X[/tex] is almost geometric in nature. Let [tex]p=\frac{11}{29}[/tex]. Then
[tex]P(X = x) = \begin{cases} p & \text{if }x = 0 \\ (1-p)p & \text{if }x = 1 \\ (1-p)^2 p & \text{if }x = 2 \\ (1-p)^3p & \text{if }x = 3 \\ (1-p)^4 & \text{if }x = 4 \\ 0 & \text{otherwise}\end{cases}[/tex]
Compute the first moment/expected value.
[tex]E(X) = \displaystyle \sum_x x\, P(X=x) \\\\ ~~~~~~~~ = 0\cdot p+1\cdot(1-p)p + 2\cdot(1-p)^2p + 3\cdot(1-p)^3p + 4\cdot(1-p)^4 \approx 1.39349[/tex]
Compute the second moment.
[tex]E(X^2) = \displaystyle \sum_x x^2\, P(X=x) \\\\ ~~~~~~~~ = 0\cdot p+1\cdot(1-p)p + 4\cdot(1-p)^2p + 9\cdot(1-p)^3p + 16\cdot(1-p)^4 \approx 4.01103[/tex]
Compute the variance.
[tex]V(X) = E(X^2) - E(X)^2 \approx 2.06921[/tex]
The standard deviation is the square root of the variance.
[tex]\sqrt{V(X)} \approx \boxed{1.43848}[/tex]
