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Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mean ((a),(c),(e),(g)) and variance ((b),(d),(f),(h)) of X. Round your answers to three decimal places (e.g. 98.765).

(a),(b)f (x) = kx2 for 0 < x < 23
(c),(d)f (x) = k(1 + 2x) for 0 < x < 20
(e),(f)f (x) = ke-x for 0 < x
(g),(h)f (x) = k where k > 0 and 110 < x < 110 + k


(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)

Respuesta :

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

[tex]\int\limits^y_0 {kx^{2} } \, =1[/tex]

where y=23

Clearing k=0.00025

a) [tex]Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5[/tex]

b)[tex]Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6[/tex]

c) The function is equal to:

f(x)=k(1+2x)

[tex]\int\limits^y_0 {k(1+2x)} \, =1[/tex]

where y=20

k=0.0024

[tex]Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3[/tex]

d) [tex]Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51[/tex]

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