the-probabilty-density-function-is-known-as-follows-f-x-0-x-other-cx-3-0-lt-x-lt-4-define-P-1-lt-x-lt-2-

Question Number 145675 by syamilkamil1 last updated on 07/Jul/21

t h e p r o b a b i l t y d e n s i t y f u n c t i o n

i s k n o w n a s f o l l o w s :

f (x) = {_{0, x o t h e r}^{{c x}^{3}, 0 < x < 4}

d e f i n e P (1 < x < 2)!

Answered by imjagoll last updated on 07/Jul/21

\int_{0}^{4} f (x) dx = 1

\Rightarrow \int_{0}^{4} {cx}^{3} dx = 1

\Rightarrow \frac{1}{4} c (4^{4}) = 1 \Rightarrow c = \frac{1}{4^{3}}

P (1 < x < 2) = \int_{1}^{2} \frac{1}{64} x^{3} dx

= \frac{1}{256} (16 - 1) = \frac{15}{256}

Answered by puissant last updated on 07/Jul/21

\int_{R} {cx}^{3} dx = 1 \Rightarrow c \int_{0}^{4} x^{3} dx = 1

\Rightarrow c {[\frac{x^{4}}{4}]}_{0}^{4} = 1 \Rightarrow 64 c = 1

c = \frac{1}{64}

P (1 < x < 2) = \frac{1}{64} \int_{1}^{2} x^{3} dx = \frac{1}{256} (16 - 1)

P (1 < x < 2) = \frac{15}{256} = 0, 059

P (1 < x < 2) = 5, 9 % . .

Answered by Dwaipayan Shikari last updated on 07/Jul/21

f (x) = {c x}^{3}

\int_{0}^{4} {c x}^{3} d x = 1 \Rightarrow c = \frac{1}{64}

P (1 < x < 2) = \frac{1}{64} \int_{1}^{2} x^{3} d x = \frac{1}{256} (2^{4} - 1) = \frac{15}{256}