Continuous Random Variables – Measures of Centre

Continuous randoms use the same measures of centre as the discrete random variables. The measures of centre that we will use are the mean median and the mode.

Mean

The mean is the expected value that will occur from the probability distribution.

$E(x)=\int\limits_{-\infty}^{\infty}{xf(x)}dx$

For most of our functions the probability density function will be equal to zero for most of the domain. For example a long jumper has zero chance of jumping 100 meters and will most likely have his results between two to eight meters.

$f(x)=0 for all but [a,b], then E(X)=\int\limits_{a}^{b}{xf(x)}dx$

Median

The median is simply the middle value of the probability distribution. Given that we have a probability function, we are able to solve the equation of the integral for the variable p for a given percentage q.

$\int\limits_{-\infty}^{p}{f(x)}dx=q$

If we let q=0.5 then we can solve for p to get a median value. It is also substitute q for other values such as q=0.9 to get the range of the whole 90^th percentile.

Also we only need to perform the integral over the range of our density function, therefore we can change the negative infinity value for the bottom end of the range of our probability function.

Mode

The mode is the most common value that occurs in the function and is not actually a measure of centre. In order to find the mode, we need to find the maximum of the probability density function.

As soon as we know we need to find a maximum we know that we need to find a derivative of the function and solve for zero. Then if the gradient is positive prior to the turning point and negative after we will have our maximum point.

Practice Exam Question

Question 8 from the 2014 exam 1

Step 1 Use our integral integration rules to integrate the function.

$\int\limits_{0}^{m}{\frac{1}{5}e^{-\frac{x}{5}}}dx=0.5$

$[-e^{-\frac{x}{5}}]_{0}^{m}=0.5$

$-e^{-\frac{m}{5}}-(-e^{-\frac{m}{5}})=0.5$

$e^{-\frac{m}{5}}=1-0.5$

Then we take logs of both sides

$log_{e}e^{-\frac{m}{5}}=log_{e}0.5$

$-\frac{m}{5}=-log_{e}2$

$m=5log_{e}2$

This is a tricky question as it relies on our understanding of continuous variables, integration and log laws. This is another good example of why we need to know all of our topics well to answer the final exam questions.

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