Let f(x) = kx2(2 − x) if 0 ≤ x ≤ 2 and f(x) = 0 if x < 0 or x > 2. (a) For what value of k is f a probability density function? k =_______ (b) For that value of k, find P X ≥ 1 . (c) For that value of k, find the mean.

Answer:(a) k = 3/4(b) 11/16(c) 1.2Step-by-step explanation:(a) for a probability density function,int(-inf,inf, f(x)) = 1i.e. -inf is the lower bound, inf is the upper bound f(x) is the function in integrationsince range of x is between 0 and 2 the equation becomesint(0,2, f(x)) = 1int(0,2, kx2(2-x)) = 1expand f(x): kx2(2-x) = 2kx2 - kx3integrate f(x) from x= 0 to 2:int(0,2, 2kx2-kx3)= k( 2x3/3 - x4/4) | (0 to 2)= k (2(8)/3 - 16/4 - 0)  (after substitute 2 and 0 into the definite integral)= k (4/3)= 4k/3for a probability density function 4k/3 = 1Hence k = 3/4(b)new function f(x) = (3/4) x2(2-x)find [tex]P\geq 1[/tex]The function has max x value of 2.[tex]P(X\geq 1) = \int\limits^a_b {\frac{3}{4} x^{2}(2-x) } \, dx[/tex]with a = 2, b = 1Through definite integral, we'll get[tex]\frac{3}{4} [ {(\frac{16}{3}-4)-(\frac{2}{3}-\frac{1}{4} )][/tex]= [tex]\frac{11}{16}[/tex](c)Mean = [tex]\int\limits^a_b {x.f(x)} \, dx[/tex] with a = 2 and b = 0 [tex]x.f(x) = \frac{3}{4}x^{2}(2-x)\\ =[tex]\int\limits^a_b {\frac{3}{2}x^{3} -\frac{3}{4} x^{4}} \, dx[/tex]= [tex]\frac{3}{4}x^{4}(1/4) - \frac{3}{4}x^{5}(1/5) | a =2, b=0[/tex]= [tex]\frac{3}{8}(16) - \frac{3}{20}(32) - 0[/tex]= 6 - 4.8= 1.2