Question

For some constant c, the random variable X has the probability density function f(x) = c x n 0 < x < 1 0 otherwise Find (a) c and

(b) P{X > x}, 0 < x < 1.

Short answer

The result is $\left(\mathrm{a}\right)\mathrm{c}=\mathrm{n}+1$

(b)$P(X>x)=1-x^{n+1}$

Step-by-step solution

Step:1 Given Information

Otherwise, the random variable X has the probability density function $f\left(x\right)=c{x}^n,0<x<10$for some constant c. Find (a) c and$\text{(b)}P\{X>x\},0<x<1\text{.}$

Step:2 Definition

A probability density function (PDF) is used in probability theory to characterize the random variable's likelihood of falling into a specific range of values rather than taking on a single value. The function illustrates the normal distribution's probability density function and how mean and deviation are calculated.

Step:3 Explanation of the solution

(a) Condition must be met by the probability density function.

${\int }_{\mathrm{ℝ}}fdx=1$

Hence, there has to be

$1={\int }_{\mathrm{ℝ}}f\left(x\right)dx={\int }_0^{1}c{x}^{n}dx={\left.c·\frac{x^{n+1}}{n+1}\right|}_0^1=\frac{c}{n+1}$

Finally, the c constant must equal n+1.

(b) Calculate the requisite probability as the integral of f over the suitable

region. That is something we have.

$P(X>x)={\int }_x^{1}f\left(s\right)ds={\int }_x^1(n+1)s^{n}ds={\left.s^{n+1}\right|}_x^1=1-x^{n+1}$