Continuous random variable

[Frank_Ma]

A continuous random variable X has probability density function: f(x) = kx 0 < x < 2
otherwise 0

a. Determine the value of k.
No idea, except that maybe K has to be zero? Because any individual probability = 0?
b. Find E(X)and Var(X).
E(x) = (a+b)/2? and Var(X) = (a+b)^2 / 12? where a = 0 and b=2
c. What is the probability that X is greater than three standard deviations above the mean?
1 - 99.6%? and then divide by 2?
d. Find the distribution function F(X) and hence the median of X.
No idea?

 

[pka]

A continuous random variable X has probability density function: f(x) = kx 0 < x < 2
otherwise 0

a. Determine the value of k.
No idea, except that maybe K has to be zero? Because any individual probability = 0?
b. Find E(X)and Var(X).
E(x) = (a+b)/2? and Var(X) = (a+b)^2 / 12? where a = 0 and b=2
c. What is the probability that X is greater than three standard deviations above the mean?
1 - 99.6%? and then divide by 2?
d. Find the distribution function F(X) and hence the median of X.
No idea?

a) Find \(k\) so that \(\displaystyle\int_0^2 {kxdx} = 1\).
Then go to your textbook/notes to see the other parts.
Do something on them and post your efforts here to receive more help.

 

[Frank_Ma]

Is there a way to do this without integration though? Have found that way of doing it online. But, im doing a stats course only and no integration was covered.

Thanks for your help!

 

[pka]

Is there a way to do this without integration though? Have found that way of doing it online. But, im doing a stats course only and no integration was covered.

What textbook are you using? The very statement of the question implies the use if calculus.
The statement that \(f\) is a probability density function means \(0\le f(x)\le1~\&~\displaystyle \int_{ - \infty }^\infty f = 1\).
Moreover, the part b & c are both integration questions. So I can't understand your no knowing caculus.

 

[Frank_Ma]

Statistics for business and economics - Anderson.
Maybe it is only possible through integration. Would not be beyond the creativity of our professor!
Thanks for your help anyway Pka.