1 Attachment(s)

Cumulative Probability & Mean (Discrete & Continuous)

Hi Members,

I am in a panic trying to work out the formulas below, I am unsure if I have posted in the correct place, my lecturer said it is basic primary statistics, *(which I can’t believe, not in my primary school) *I have to solve a similar formula for my “*Simulation of Multi-media Networks*” exam at university.

The lecturer went over it all at breakneck speed and I could not grasp it, I am a mature student and it has been a long time since I did maths and as far as statistics are concerned I have only learned to do quartiles, mean & median. I know it is a big ask but I was hoping someone could help break down the formulas and explain how the process the lecturer went through to get her answers. *(I will be provided with the formula in the exam, which will be no good unless I know how to use it) *I have copied the details below as they are from handouts.

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**Cumalative Probability Distribution:**

Cumulative probabilities from a (discrete) probability distribution..

$\displaystyle

\begin{tabular}{|c|c|c|c|}

\hline

$x$&$0$&$1$&$2$\\

\hline

P(x)&0.2&0.5&0.3\\

\hline

\sum P(x)&0.2&0.7&1\\

\hline

\end{tabular}

$

Note: I can kind of see how row $\displaystyle \sum P(x)$ is worked out, but I can't figure out how row $\displaystyle P(x)$ is worked out at all.

**Mean or Expected Value:**

Discrete:

$\displaystyle Mean=\bar{x}=E(x)=\sum_{i=1}^nx_iP(x_i)$

e.g.

$\displaystyle

\begin{tabular}{|c|c|c|c|}

\hline

$x$&$0$&$1$&$2$\\

\hline

P(x)&0.2&0.5&0.3\\

\hline

\end{tabular}

$

**Mean=0x0.2+1x0.5+2x0.3=1.1**

Note: I can of see the correlation of the answer to the table e.g. $\displaystyle x$ multiplied by $\displaystyle P(x)$ in each column then added together givesthe answer **1.1 **but on how this was worked out using the formula I am totally lost.

For the continuous variate:

$\displaystyle Mean=\bar{x}=E(x)=\int_{+\infty}^{+\infty}xf(x)dx$

e.g

$\displaystyle f(x)=2x$Attachment 19406$\displaystyle 0\leq x \leq1$

$\displaystyle

E(x)=\int_0^12x^2dx=\left[ \frac{2x^3}{3} \right]_0^1=\frac{2}{3}

$

Note: On the continuous variate I am also totaly lost.

Any help anyone has to offer to better my understanding of this formula would be gratefuly appreciated.

Colin