Question

A random variable gives measurements X between 0and1 with a probability function F(x)= 12x^3 -21x^2+10x. 0<=x<=1
find the number k such that p(x<=k)=1/2​

Answer 1

Answer:

Himanshu tu hi himanshi khurana ko janta hai ky??

Answer 2

Given : A random variable gives measurements X between 0and1 with a probability function F(x)= 12x^3 -21x^2+10x. 0<=x<=1

To find : the number k such that p(x<=k)=1/2​

Solution:

f(x)= 12x³ -21x²+10x.    0<=x<=1

$\int\limits^1_0 {(12x^3-21x^2+10x)} \, dx$

=  3x⁴ - 7x³ + 5x²]$]_0^1$

=  3 - 7  + 5

= 1

number k such that p(x<=k)=1/2​

$\int\limits^k_0 {(12x^3-21x^2+10x)} \, dx =\dfrac{1}{2}$

=>  3x⁴ - 7x³ + 5x²]$]_0^k$  = 1/2

=> 3k⁴  - 7k³  + 5k²   = 1/2

=> 6k⁴ - 14k³ + 10k² = 1

=> 6k⁴ - 14k³ + 10k² - 1 = 0

on solving k ≈0.45176

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