Question

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 1

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 $P\geq 1$

The function has max x value of 2.

$P(X\geq 1) = \int\limits^a_b {\frac{3}{4} x^{2}(2-x) } \, dx$

with a = 2, b = 1

Through definite integral, we'll get

$\frac{3}{4} [ {(\frac{16}{3}-4)-(\frac{2}{3}-\frac{1}{4} )]$

= $\frac{11}{16}$

(c)

Mean = $\int\limits^a_b {x.f(x)} \, dx$ with a = 2 and b = 0

$x.f(x) = \frac{3}{4}x^{2}(2-x)\\ =[tex]\int\limits^a_b {\frac{3}{2}x^{3} -\frac{3}{4} x^{4}} \, dx$

= $\frac{3}{4}x^{4}(1/4) - \frac{3}{4}x^{5}(1/5) | a =2, b=0$

= $\frac{3}{8}(16) - \frac{3}{20}(32) - 0$

= 6 - 4.8

= 1.2