continuous random variable find F: f(x) = {C((1/2)x + 5, for x E [2,3] and 0, for x N

[tweetbird]

Hi, i need a help with continuos random variable. I have this exercise but I can’t solve it. I solved half of it but don’t know how to finish it.

The exercise is:

We have function f(x) = {C((1/2)x + 5, for x E [2,3] and 0, for x Not E [2,3]. determine the constant C for function f to be the probability density function of a continuous random variable X. If F (x) is the distribution function of the probability of random variables X, calculate F (1), F (2.5), F (4).

I know how to get C but I do not know how to calculate F(1),F(2.5),F(4).

When i integrate C I get:
C= 4/25.

Can anyone tell me how and what to write to get all the Fs.

Firstly I wrote:

F(1) = 4/25 * (1/2 * 1 + 5) the value that I put inside the parenthesis is the value from the beginning from C (1/2x +5)

but my professor told me that is not right. Can anyone tell me what should I write there and how to get it

< link to image of young girls in sexual poses removed >

 

[tweetbird]

photo

Here is the photo how I tried to solve it and what was wrong

 

[Harry_the_cat]

Hi, i need a help with continuos random variable. I have this exercise but I can’t solve it. I solved half of it but don’t know how to finish it.

The exercise is:

We have function f(x) = {C((1/2)x + 5, for x E [2,3] and 0, for x Not E [2,3]. determine the constant C for function f to be the probability density function of a continuous random variable X. If F (x) is the distribution function of the probability of random variables X, calculate F (1), F (2.5), F (4).

I know how to get C but I do not know how to calculate F(1),F(2.5),F(4).

When i integrate C I get:
C= 4/25.

Can anyone tell me how and what to write to get all the Fs.

Firstly I wrote:

F(1) = 4/25 * (1/2 * 1 + 5) the value that I put inside the parenthesis is the value from the beginning from C (1/2x +5)

but my professor told me that is not right. Can anyone tell me what should I write there and how to get it

the photo: https://imgur.com/a/hUQyJx0

You mention f(x) and F(x), they are not the same thing.
You say f(x) is the probability density function. Then you say that F(x) is the distribution function. Do you mean that F(x) is the cumulative distribution function?

 

[HallsofIvy]

Assuming that "f(x)" is a probability density function and that "F(x)" is its cumulative probability function, \(\displaystyle F(x)= \int_{-\infty}^x f(t)dt\). Here, f(x) is non- zero only on the interval [2, 3] so the lower bound on that integral is "2" rather than \(\displaystyle -\infty\).

One problem I have is that your function, f, is unclear: you have "C((1/2)x+ 5" so there is a missing ")". Do you mean C(1/2)x+ 5 or C((1/2)x+ 5)?

If it is the former then we must have \(\displaystyle \frac{C}{2}\int_2^3 xdx+ 5= \frac{C}{2}(5/2)+ 5= 5C/4+ 5= 1\) so C= -16/5. If it is the latter then we must have \(\displaystyle C\int_2^4 \frac{x}{2}+ 5 dx= C\left(\frac{x^2}{4}+ 5x\right)_2^3= C\left(\frac{9}{4}+ 15- 1- 10\right)= \frac{25}{4}C= 1 so \(\displaystyle C= \frac{4}{25}\). Which did you get?

In any case, F(x) must be 0 for x< 2, \(\displaystyle \int_2^x f(t)dt\) for \(\displaystyle 2\le x\le 3\), and 1 for x> 3.\)