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The variance of Y, denoted by Var(Y), is defined as the expected value of the squared deviation from the mean and can be calculated as follows:
Var(Y) = E((Y - E(Y))^2) = E(Y^2) - (E(Y))^2
= ∫y^2 * g(y) dy - (E(Y))^2
= ∫y^2 * 4y^3 dy - (E(Y))^2
= ∫4y^5 dy - (E(Y))^2
= y^6/6 | from 0 to 1
= (1^6)/6 - (0^6)/6
= 1/6 - 0
= 1/6
So the variance of Y is 1/6.
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- YOUR TURN 1 Repeat Example 1 for the probability density function f(t)=83x3 on [1,2]. EXAMPLE 1 Expected Value and Variance Find the expected value and variance of the random variable x with probability density function defined by f(x)=(3/26)x2on[1,3].If a dealer’s profit, in units of $5000, on a new automobile can be looked upon as a random variable X having the density function Find the variance of X.What is the variance of a continuous probability distribution function with the probability density function f(x) = 4(x-1)^2 for 0 < x < 2?