Question

suppose the density function of X is defined as, f(x)= 3x^2, 0<=x<=1 Calculate variance of X.​

Answer 1

Answer:

Variance of X is 3/80

Step-by-step explanation:

Density function of X is defined as , f(x) = 3x² , 0≤x≤1

Now, E(X) = ∫xf(x) dx ,lower limit of the integration is -∞ and upper limit is ∞

=> E(X) = ∫3x³ dx , lower limit is 0 and upper limit is 1

=> E(X) = 3/4 [x^4] ,lower limit is 0 and upper limit is 1

=> E(X) = 3/4 [1-0] = 3/4

Now, E(X²) = ∫x²f(x) dx ,lower limit of the integration is -∞ and upper limit is ∞

=> E(X²) = ∫3x^4 dx , lower limit is 0 and upper limit is 1

=> E(X²) = 3/5 [x^5] ,lower limit is 0 and upper limit is 1

=> E(X²) = 3/5 [1-0] = 3/5

Now , we know that, Var(X) = E(X²) - (E(X))²

=> Var(X) = 3/5 - (3/4)² = 3/5 - 9/16 = 3/80

Answer 2

Answer:

2.2

Step-by-step explanation: