Answer:
The standard deviation of given probability distribution is 0.8292
Step-by-step explanation:
We are given the following in the question:
q: 0 1 2
Probability: 0.25 0.25 0.50
Formula:
[tex]E(q) = \displaystyle\sum q_ip(q_i)\\=0(0.25) + 1(0.25) + 2(0.50) = 1.25[/tex]
[tex]E(q^2)= \displaystyle\sum q_i^2p(q_i)\\=0^2(0.25) + 1^2(0.25) + 2^2(0.50) = 2.25[/tex]
Variance =
[tex]\sigma^2 = E(q^2) = (E(q))^2\\= 2.25- (1.25)^2\\=0.6875\\\sigma = \sqrt{0.6875} = 0.8292[/tex]
Thus, the standard deviation of given probability distribution is 0.8292
