4 years ago

I need help please no one helps me

Mathematics

2 answers:

Natasha2012 [34]4 years ago

6 0

Answer:

it's a

Step-by-step explanation:

andriy [413]4 years ago

3 0

Answer:

answer is a.

Step-by-step explanation:

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For the following distribution,

Musya8 [376]

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Step-by-step explanation:

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X: 0 1 2 3

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$E(X) = \displaystyle\sum x_iP(x_i)\\\\E(X) = 0(0.027) + 1(0.189) + 2(0.441) + 3(0.343)\\E(X) = 2.1\\\\E(X^2) = \displaystyle\sum x_i^2P(x_i)\\\\E(X^2) = 0^2(0.027) + 1^2(0.189) + 2^2(0.441) + 3^2(0.343)\\E(X^2) = 5.04\\\\Var(X) = E(X^2) -[E(X)]^2\\Var(X) = 5.04 - (2.1)^2\\Var(X) = 0.63$

The variance of given distribution is 0.63

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