Answer:
(a) k = 3/4
(b) 11/16
(c) 1.2
Step-by-step explanation:
(a)
for a probability density function,
int(-inf,inf, f(x)) = 1
i.e. -inf is the lower bound, inf is the upper bound f(x) is the function in integration
since range of x is between 0 and 2 the equation becomes
int(0,2, f(x)) = 1
int(0,2, kx2(2-x)) = 1
expand f(x): kx2(2-x) = 2kx2 - kx3
integrate 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/3
for a probability density function 4k/3 = 1
Hence 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 = 1
Through 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
