Math  /  Algebra

QuestionAssume XX has a normal distribution N(9,52)N\left(9,5^{2}\right). Find E(5X4)2E(5 X-4)^{2}
Answer: \square

Studdy Solution
Now, substitute these values into the formula for E(Y2) E(Y^2) :
E(Y2)=Var(Y)+[E(Y)]2=625+412 E(Y^2) = \text{Var}(Y) + [E(Y)]^2 = 625 + 41^2
Calculate 412 41^2 :
412=1681 41^2 = 1681
Therefore:
E(Y2)=625+1681=2306 E(Y^2) = 625 + 1681 = 2306
The expected value E(5X4)2 E(5X - 4)^2 is 2306 \boxed{2306} .

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