Probability Mass Function And Probability Density Function

A random variable $\mathrm{X}$ has the density function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{cc}\frac{1}{4}& -2<\mathrm{x}<2\\ 0& \mathrm{elsewhere}\end{array}\right.$ .Then, find $\mathrm{P}(\mathrm{X}>1)$.(Use concept of Probability Density Function)

A random variable $\mathrm{X}$ has the density function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{cc}\frac{1}{4}& -2<\mathrm{x}<2\\ 0& \mathrm{else} \mathrm{where}\end{array}\right.$ .Then find $\mathrm{P}(-1<\mathrm{X}<2)$(Use concept of Probability Density Function).

Verify the following is a p.d.f $\mathrm{f}\left(\mathrm{x}\right)=5{\mathrm{x}}^4, 0<\mathrm{x}<1$? Explain.

Distinguish between probability mass function and probability density function.

$\mathrm{X}$ is a discrete random variable which takes the value $0,1,2$ and $\mathrm{P}(\mathrm{X}=0)=\frac{144}{169}; \mathrm{P}(\mathrm{X}=1)=\frac{1}{169}$ and $\mathrm{P}(\mathrm{X}=2)=\mathrm{K},$ then the value of $\mathrm{K}$ is 

A random variable $\mathrm{X}$ has the following probability distribution.

Then, $\mathrm{P}(1\le \mathrm{X}\le 4)$ is

$f\left(x\right)=\left\{\begin{array}{cl}k{x}^2& 0\le x\le 3\\ 0& \text{ elsewhere }\end{array}\right.$is a probability density function then the value of $k$ is