# Thread: Cumulative Probability & Mean (Discrete & Continuous)

1. ## 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$$\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

2. For your first question, the values of P(X) at x = 0, 1, and 2 are given to you from which you find $\displaystyle \sum P(x)$

To find the mean (of a discrete distribution), which is also called the Expected Value of x:

$\displaystyle \sum x \times P(X=x)$

As stated by your question, this discrete random variable X takes the values 0, 1 and 2. Then the Mean or the Expected Value is calculated by multiplying each x value by its probability and summing them up.

$\displaystyle E[X] = 0P(X=0)+1P(X=1)+2P(X=2)= 0(0.2)+1(0.5)+2(0.3)$

The expected value or the mean of a continuous random variable, X, that has a value between a and b(In this case, 0 and 1) is computed by integrating x times its probability density function (p.d.f.) over the interval [a,b]. Your pdf here is f(x)=2x over the interval[0,1]:

$\displaystyle \displaystyle{E(X) = \int_0^1 x \times f(x) dx = \int_0^1 x \times 2x \mbox{dx} = \int_0^1 2x^2 \mbox{dx}}$

3. Thanks very much harish21.