Respuesta :
We have to find the standard deviation of this probability distribution.
We have to start by calculating the mean of this distribution.
It can be calculated as:
[tex]\begin{gathered} \bar{x}=\sum^nP(x_i)\cdot x_i \\ \bar{x}=0.05\cdot0+0.15\cdot1+0.1\cdot2+0.7\cdot3 \\ \bar{x}=0+0.15+0.2+2.1 \\ \end{gathered}[/tex]We now can calculate the standard deviation using the last result as:
[tex]\begin{gathered} \sigma=\sqrt{\sum_^P(x_i)\cdot(x_i-\bar{x})^2} \\ \sigma=\sqrt{0.05(0-2.1)^2+0.15(1-2.1)^2+0.1(2-2.1)^2+0.7(3-2.1)^2} \\ \sigma=\sqrt{0.05(-2.1)^2+0.15(-1.1)^2+0.1(-0.1)^2+0.7(0.9)^2} \\ \sigma=\sqrt{0.05(4.41)+0.15(1.21)+0.1(0.01)+0.7(0.81)} \\ \sigma=\sqrt{0.2205+0.1815+0.001+0.567} \\ \sigma=\sqrt{0.97} \\ \sigma\approx0.98 \end{gathered}[/tex]Answer: the standard deviation is approximately 0.98.
