Distribution Function F

Hello...!!!I need some help...!!!
Let the distribution function F of a random variable X given in the following attachment. Calculate the following:
P(X=-1), P(X<0), P(X<=0), P(X=1), P(X>5), P(X>=5), P(3<=X<=4).

I think that these are the answers(X<0)=F(0-)=0.1, P(X<=0)=F(0)=0.2, P(3<=X<=4)=F(4)-F(3)=0.8-0.8=0, P(X>5)=P(X>=5)=0, P(X=-1)=F(-1+)-F(-1-)=0.1-0=0.1, P(X=1)=F(1+)-F(1-)=0.3-0.3=0,but I am not sure...
I hope you can help me....!!!!!

mathmari said:

Hello...!!!I need some help...!!!
Let the distribution function F of a random variable X given in the following attachment.

Click to expand...

??? There is no attachment.
Would it really be that difficult to type it in here?

Calculate the following:

Click to expand...

P(X=-1), P(X<0), P(X<=0), P(X=1), P(X>5), P(X>=5), P(3<=X<=4).

I think that these are the answers(X<0)=F(0-)=0.1, P(X<=0)=F(0)=0.2, P(3<=X<=4)=F(4)-F(3)=0.8-0.8=0, P(X>5)=P(X>=5)=0, P(X=-1)=F(-1+)-F(-1-)=0.1-0=0.1, P(X=1)=F(1+)-F(1-)=0.3-0.3=0,but I am not sure...
I hope you can help me....!!!!!

Click to expand...

Oh sorry!!!!!!!!!!!!!!!!!! :? Here is the attachment:

mathmari said:

Hello...!!!I need some help...!!!
Let the distribution function F of a random variable X given in the following attachment. Calculate the following:
P(X=-1), P(X<0), P(X<=0), P(X=1), P(X>5), P(X>=5), P(3<=X<=4).

I think that these are the answers(X<0)=F(0-)=0.1, P(X<=0)=F(0)=0.2, P(3<=X<=4)=F(4)-F(3)=0.8-0.8=0, P(X>5)=P(X>=5)=0, P(X=-1)=F(-1+)-F(-1-)=0.1-0=0.1, P(X=1)=F(1+)-F(1-)=0.3-0.3=0,but I am not sure...
I hope you can help me....!!!!!

Click to expand...

You certainly have the correct ideas.

\(\displaystyle P(X\ge 5)=1-F(5-)\)

\(\displaystyle P(3\le X\le 4)=F(4)-F(3-)\)

mathmari said:

So is P(X>5)=1-P(X<=5)=1-F(5)=1-1=0??? and are the following right??? P(0<X<=3)=F(3)-F(0+)=0.8-0.2=0.6, P(0<X<3)=F(3-)-F(0+)=0.6-0.2=0.4, P(0<=X<=3)=F(3)-F(0)=0.8-0.2=0.6, P(1<X<=2)=F(2)-F(1+)=0.3-0.3=0, P(1<=X<=2)=F(2)-F(1)=0

Click to expand...

Look at it this way:
\(\displaystyle P(X\le 3)\) is the accumulation on \(\displaystyle (-\infty,3]\) which is \(\displaystyle F(3)\).

Thus \(\displaystyle P(0<X\le 3)\) is the accumulation on \(\displaystyle (-\infty,3]\setminus (-\infty, 0]\) which is \(\displaystyle F(3)-F(0)\).

Think in terms of complements.

So \(\displaystyle P(0<X< 3)\) is the accumulation on \(\displaystyle (-\infty,3)\setminus (-\infty,0]\) which is \(\displaystyle F(3-)-F(0)\)

\(\displaystyle \begin{array}{*{20}{c}} {1.}&{P(X \leqslant a) = F(a)} \\ {2.}&{P(X < a) = F(a^- )} \\ {3.}&{P(a < X \leqslant b) = F(b) - F(a)} \\ {4.}&{P(X = a) = F(a) - F(a^- )} \\ {5.}&{P(a < X < b) = F(b^- ) - F(a)} \\ {6.}&{P(a \leqslant X \leqslant b) = F(b) - F(a^- )} \\ {7.}&{P(a \leqslant X < b) = F(b^- ) - F(a^- )} \\ {8.}&{P(a < X) = 1 - F(a)} \\ {9.}&{P(a \leqslant X) = 1 - F(a^- )} \end{array}\)

Thank you very much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!