# Estimation of probability of the distribution of a continuous random variable [closed]

I am trying to find the probability distribution of a continuous random variable X which is defined by 𝑃(X=x)=𝑘(2−x)(x+1),where 0≤=x≤2

Can anybody help me to find x and mean expected value E(x) ?

• First of all you should evaluate the value of $k$. Use that $\int\limits_{0}^2 f(x) \, dx =1$ Commented Dec 26, 2020 at 15:36
• I suppose that $P(X=x)$ in the above really should be $f_X(x)$ (the probability density function of $X$), as for a continuous random variable we have that $P(X=x)=0$. Commented Dec 26, 2020 at 15:45

You can't find $$x$$, the question doesn't make sense and I suspect you meant find $$k$$ so I will proceed with that A continuous random variable must satisfy the property $$\int f(x) dx=1$$ or in other words: the sum of all probability must equal to $$1$$.

To find $$k$$ we need $$k\times\int_0^2(2-x)(x+1)dx=1$$, we have $$\int_0^2(2-x)(x+1)dx=\frac{10}{3}$$ (this was done quickly on WolframAlpha as the goal here isn't to teach you integration) so we have $$k\times \frac{10}{3}=1$$ which gives us $$k=\frac{3}{10}$$

For your second question, we have $$E(X)=\int xf(x)$$ so to find the expected value you should evaluate $$\frac{3}{10}\int_0^2x(2-x)(x+1) dx$$ which gives $$\frac{4}{5}$$

• @projectilemotion yes, thank you for pointing it out, op asked as similar question about a discrete random variable like an hour ago so I just copied the initial sentence and forgot this part Commented Dec 26, 2020 at 15:48

Your question holds a couple of mistakes. First, if you are dealing with continuous random variable then, for any x the probability P(X=x) is equal to 0 by definition. Secondly, the task to find $$x$$ makes no sense, since $$x$$ is a name of an argument here.

I would assume that by P(X=x) you mean a probability density function (PDF) $$f_X(x)$$, which is also defined as 0 everywhere outside $$x\in [0,2]$$, and will build my answer based on that information.

1. Find $$k$$. For a PDF following conditions hold by definition
• $$f_X(x)\geq0$$ for all $$x$$ (clearly holds);
• $$\int_{-\infty}^{\infty} f_X(x)dx=1$$

Calculating the integral above gives us the equation $$\frac{10}{3}\cdot k=1$$, thus $$k=\frac{3}{10}$$.

1. calculate $$E[X]$$. By definition, $$E[X]=\int_{-\infty}^{\infty} xf_X(x)dx = \int_{0}^{2} x\cdot \frac{3}{10}(2-x)(x+1)dx=\frac45$$