Answer:
Option B) 0.63
Step-by-step explanation:
We are given the following in the question:
X: 0 1 2 3
P(x): 0.027 0.189 0.441 0.343
We have to find the variance for the given discrete probability distribution.
![E(X) = \displaystyle\sum x_iP(x_i)\\\\E(X) = 0(0.027) + 1(0.189) + 2(0.441) + 3(0.343)\\E(X) = 2.1\\\\E(X^2) = \displaystyle\sum x_i^2P(x_i)\\\\E(X^2) = 0^2(0.027) + 1^2(0.189) + 2^2(0.441) + 3^2(0.343)\\E(X^2) = 5.04\\\\Var(X) = E(X^2) -[E(X)]^2\\Var(X) = 5.04 - (2.1)^2\\Var(X) = 0.63](https://tex.z-dn.net/?f=E%28X%29%20%3D%20%5Cdisplaystyle%5Csum%20x_iP%28x_i%29%5C%5C%5C%5CE%28X%29%20%3D%200%280.027%29%20%2B%201%280.189%29%20%2B%202%280.441%29%20%2B%203%280.343%29%5C%5CE%28X%29%20%3D%202.1%5C%5C%5C%5CE%28X%5E2%29%20%3D%20%5Cdisplaystyle%5Csum%20x_i%5E2P%28x_i%29%5C%5C%5C%5CE%28X%5E2%29%20%3D%200%5E2%280.027%29%20%2B%201%5E2%280.189%29%20%2B%202%5E2%280.441%29%20%2B%203%5E2%280.343%29%5C%5CE%28X%5E2%29%20%3D%205.04%5C%5C%5C%5CVar%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%5C%5CVar%28X%29%20%3D%205.04%20-%20%282.1%29%5E2%5C%5CVar%28X%29%20%3D%200.63)
The variance of given distribution is 0.63
Answer:
in a two column proof, each <u>of </u> is in the left and each <u>of</u><u> </u> is on the right !!
Point. A point is a location, yes. It can be placed anywhere on the graph.
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