SOLUTION: (a) Find the CDF of X where X is a continuous random variable with pdf f(x) = 1/3 for the interval 0 to 1 and 2/3 for the interval 1 to 2. (b) Find the median of random variable

Question 349491: (a) Find the CDF of X where X is a continuous random variable with pdf f(x) = 1/3 for the interval 0 to 1 and 2/3 for the interval 1 to 2.
(b) Find the median of random variable X.
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
a) Integrate to find the CDF.
PDF=

PDF=

Integrating,
CDF=

CDF=

Solve for the constants using conditions,

.
.
.

.
.
.
CDF=

CDF=

.
.
.
b) Median=1

RELATED QUESTIONS
A random variable X has pdf f(x) = (3/4)(1-x^2) from -1 < x < 1. 1. Find the cdf... (answered by Boreal,ikleyn)

A discrete random variable X has a pdf of the form f(x) = c(8 - x) for x = 0, 1, 2, 3, 4, (answered by Edwin McCravy)

A random variable X has a CDF such that x/2 0 < x ≤ 1 F(x) = x-1/2 (answered by Edwin McCravy,math_tutor2020)

A discrete random variable has pdf f(x). (a) If f(x) = k(1/2)^x for x= 1, 2, 3, and zero (answered by math_tutor2020)

A random variable X has the pdf x^2 if 0 < x ≤ 1 f(x) = 2/3 if 1 < x... (answered by Edwin McCravy)

Let X be a continuous random variable with density function f(x) = { theta*x +... (answered by math_tutor2020)

Please help me with this problem. I'm not really good at probability and statistics. A... (answered by stanbon)

If the probability density function of a continuous random variable X is f(x)=0.5 x for... (answered by greenestamps)

A random variable X has pdf = 1 - x/2 in interval (0, 2). What is the expected value of... (answered by CPhill)