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.

(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

The function has max x value of 2.

with a = 2, b = 1

Through definite integral, we'll get

=

(c)

Mean = with a = 2 and b = 0

=

=

= 6 - 4.8

= 1.2

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.