Parameters of Continuous Random Variable

Example

A continuous random variable \(X\) has probability density function defined as: \[f(x) = \begin{cases} \frac{x}{4}, \quad 0 \leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases}\] Calculate the mean.

Solution

We calculate the mean with the formula: \[\mu = \int_{-\infty}^{+\infty} x.f(x)dx\] Since the pdf is defined as: \[f(x) = \begin{cases} \frac{x}{4}, \quad 0 \leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases}\] The formula becomes: \[\mu = \int_0^2 x.\frac{x}{4}dx\] That's: \[\begin{aligned} \mu &= \int_0^2 \frac{x^2}{4}dx \\ & = \frac{1}{4}\int_0^2 x^2dx \\ & = \frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^2 \\ & = \frac{1}{4} \times \frac{1}{3}\begin{bmatrix}x^3\end{bmatrix}_0^2 \\ & = \frac{1}{12}\begin{bmatrix}x^3\end{bmatrix}_0^2 \\ & = \frac{1}{12}\begin{bmatrix}2^3 - 0^3\end{bmatrix} \\ & = \frac{1}{12}\begin{bmatrix}8\end{bmatrix} \\ & = \frac{8}{12} \\ & = \frac{2}{3} \\ \mu & = 0.667 \end{aligned}\] This result tells us that if we were to repeat this experiment a large number of times (hundreds, thousands, ... ) then, on average, the value of \(X\) would be \(\mu = \frac{2}{3}\).

The time, in seconds, it takes to re-heat a cup of coffee can be modelled by the continuous random variable \(X\), with probability density function: \[f(x) = \begin{cases} \frac{3}{8}x^2, \quad 0 \leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases} \] Calculate the mean amount of time it takes to re-heat a cup of coffee.

Example

The time taken, \(X\) in minutes, for certain bacteria to split into \(2\) distinct bacteria, is believed to follow a continuous probability distribution, with probability density function defined as: \[f(x) = \begin{cases} \frac{3}{8}x^2, \quad 0 \leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases}\] Assuming this model is correct, calculate the median time it takes for a bacteria to split in \(2\).

Solution

Remembering that the median is the value \(m\) such that: \[\int_{-\infty}^mf(x)dx = \frac{1}{2}\] and that the probability density function is defined as: \[f(x) = \begin{cases} \frac{3}{8}x^2, \quad 0 \leq x \leq 2 \\ 0, \quad \text{elsewhere} \end{cases}\] We find an expression for \(\int_{-\infty}^mf(x)dx\): \[\begin{aligned} \int_{-\infty}^mf(x)dx & = \int_{-\infty}^m\frac{3}{8}x^2dx \\ & = \int_0^m\frac{3}{8}x^2dx \\ & = \frac{3}{8}\int_0^mx^2dx \\ & = \frac{3}{8}\begin{bmatrix} \frac{x^3}{3}\end{bmatrix}_0^m \\ & = \frac{3}{8}\times \frac{1}{3}\begin{bmatrix} x^3\end{bmatrix}_0^m \\ & = \frac{1}{8}\begin{bmatrix} x^3\end{bmatrix}_0^m \\ & = \frac{1}{8}\begin{bmatrix} m^3-0^3\end{bmatrix} \\ \int_{-\infty}^mf(x)dx & = \frac{m^3}{8} \end{aligned}\] Since \(\int_{-\infty}^mf(x)dx = \frac{1}{2}\), we solve: \[\frac{m^3}{8} = \frac{1}{2}\] That's: \[ m^3= \frac{8}{2}\]

so \[ m^3= 4\] Finally we can state the median value, \(m\), of \(X\): \[m = \sqrt[3]{4} = 1.59\] The median value is \(1.59\), rounded to \(3\) significant figures.

This result is illustrated with the curve shown here.

The area enclosed by the curve and the \(x\)-axis, between \(x=0\) and \(x=\sqrt[3]{4}\) equals to \(0.5\).
Similarly, the area enclosed by the curve and the \(x\)-axis, between \(x=\sqrt[3]{4}\) and \(x=2\) equals to \(0.5\).

This median value tells us that, when carrying-out this experiment, there is a \(50\%\) chance that the continuous random variable be less than \(\sqrt[3]{4}\) (\(\approx 1.59 \)). There is, consequently, a \(50\%\) that it be greater that \(\sqrt[3]{4}\).

In this question's context this median value tells us that there is a \(50\%\) chance that the bacteria split into \(2\) bacteria in less than \(1.59\) minutes. Likewise, there is a \(50\%\) chance that the bacteria split into \(2\) bacteria in more than \(1.59\) minutes.

Example

The weight (or mass) \(X\), in kg, of newborn babies can, roughly, be approximated by the continuous probability distribution defined as:

\[f(x) = \begin{cases} \frac{\pi}{4}sin\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix}, \quad 0\leq x \leq \pi \\ 0, \quad \text{elsewhere} \end{cases} \] Find the mode (modal value) of the newborn's weight.

Note: a more accurate model will be seen when we learn about normal distributions.

Solution

To find the mode we look for any local maximum on the non-zero portion of the probability density function's curve.

To do this we solve: \[f'(x) = 0\] Since the non-zero portion of \(f(x)\) is \(\frac{\pi}{4}sin\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix}\), for \(2\leq x \leq 4\) we differentiation: \[f(x) = \frac{\pi}{4}sin\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix}\] That's: \[f'(x) = \frac{\pi^2}{8}cos\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix}\] The solution(s) to \(f'(x)=0\) are the solutions to: \[\frac{\pi^2}{8}cos\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix} = 0\] We solve here: \[\begin{aligned} &\frac{\pi^2}{8}cos\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix} = 0 \\ & cos\begin{pmatrix}\frac{\pi (x-2)}{2} \end{pmatrix} = 0\\ & \frac{\pi (x-2)}{2} = \frac{\pi}{2} + k\pi, \quad k\in \mathbb{Z} \\ & x - 2 = 1 + 2k, \quad k\in \mathbb{Z} \\ & x = 3 + 2k, \quad k\in \mathbb{Z} \end{aligned}\] The only solution that lies within the interval \(2\leq x \leq 4\) is \(x=3\).

So the solution to the equation \(f'(x)=0\) is: \[x = 3\] This is the mode of the continuous random variable: \[\text{Mode} = 3\] This result tells us that \(3\) is the value most likely to lie within the range of the outcome of this experiment.
Given this example's context, a newborn's weight is more likely to lie within an interval containing \(3\)kg than any other.

Example

The time it takes to open an app, on a smartphone, varies from one device to another.
The time taken \(X\), in seconds, to open/launch the app "Monkey Heaven" has probability density function defined as: \[f(x) = \begin{cases} \frac{(x-3)^2}{9}, \quad 0 \leq x \leq 3 \\ 0, \quad \text{elsewhere} \end{cases}\]

1. Find the mean value of \(X\)..
2. Find the variance.standard deviation.
3. Find the standard deviation.

Solution

1. We calculate the mean using the definition: \[\mu = \int_{-\infty}^{+\infty}x.f(x)dx\] Since \(f(x)\) is defined as: \[f(x) = \begin{cases} \frac{(x-3)^2}{9}, \quad 0 \leq x \leq 3 \\ 0, \quad \text{elsewhere} \end{cases}\] This becomes: \[\mu = \int_0^3 x.\frac{(x-3)^2}{9} dx \] We now evaluate this integral to find the mean: \[\begin{aligned} \mu & = \int_0^3 x.\frac{(x-3)^2}{9} dx \\ & = \frac{1}{9}\int_0^3 x.(x-3)^2 dx \\ & = \frac{1}{9}\int_0^3 x.(x^2-6x+9) dx \\ & = \frac{1}{9}\int_0^3 (x^3-6x^2+9x) dx \\ & = \frac{1}{9}\begin{bmatrix} \frac{x^4}{4} - \frac{6}{3}x^3 + \frac{9}{2}x^2 \end{bmatrix}_0^3 \\ & = \frac{1}{9}\begin{bmatrix} \frac{x^4}{4} - 2x^3 + \frac{9}{2}x^2 \end{bmatrix}_0^3 \\ & = \frac{1}{9}\begin{bmatrix} \begin{pmatrix} \frac{3^4}{4} - 2\times 3^3 + \frac{9}{2}\times 3^2 \end{pmatrix} - 0\end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{81}{4} - 2\times 27 + \frac{9}{2}\times 9 \end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{81}{4} - 54 + \frac{81}{2} \end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{81}{4} - \frac{216}{4} + \frac{162}{4} \end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{27}{4}\end{bmatrix} \\ & = \frac{27}{9\times 4} \\ & = \frac{3}{4} \\ \mu & = 0.75 \end{aligned}\] The mean is \(\mu = 0.75\), this tells us that on average (after a large number of trials) we can expect the average time it takes (regardless of the device) to launch/open the app to be \(0.75\) seconds.

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2. To calculate the variance, we use the definition: \[Var\begin{pmatrix}X\end{pmatrix} = E\begin{pmatrix}X^2\end{pmatrix} - \mu^2\] We already know the value of \(\mu \).
   We now calculate \(E\begin{pmatrix}X^2\end{pmatrix}\): \[\begin{aligned} E\begin{pmatrix}X^2\end{pmatrix} &= \int_{-\infty}^{+\infty}x^2.f(x)dx \\ &= \int_{-\infty}^{+\infty} x^2.\frac{(x-3)^2}{9}dx \\ &= \int_0^3 x^2.\frac{(x-3)^2}{9}dx \\ &= \frac{1}{9}\int_0^3 x^2.(x-3)^2dx \\ &= \frac{1}{9}\int_0^3 x^2.(x^2-6x+9)dx \\ &= \frac{1}{9}\int_0^3 (x^4-6x^3+9x^2)dx \\ &= \frac{1}{9}\begin{bmatrix} \frac{x^5}{5}-\frac{6}{4}x^4+\frac{9}{3}x^3 \end{bmatrix}_0^3 \\ & = \frac{1}{9}\begin{bmatrix} \frac{x^5}{5}-\frac{3}{2}x^4+3x^3 \end{bmatrix}_0^3 \\ & = \frac{1}{9}\begin{bmatrix} \begin{pmatrix} \frac{3^5}{5}-\frac{3}{2}\times 3^4+3\times 3^3 \end{pmatrix} - 0\end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \begin{pmatrix} \frac{243}{5}-\frac{243}{2}+81 \end{pmatrix} - 0\end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{486}{10} - \frac{1215}{10} + \frac{810}{10}\end{bmatrix} \\ & = \frac{1}{9}\begin{bmatrix} \frac{81}{10} \end{bmatrix} \\ & = \frac{81}{9\times 10} \\ & = \frac{9}{10} \\ E\begin{pmatrix}X^2\end{pmatrix} & = 0.9 \end{aligned}\] We can now go back to the definition of variance and calculate: \[\begin{aligned} Var\begin{pmatrix}X \end{pmatrix} &= E\begin{pmatrix}X^2\end{pmatrix} - \mu^2 \\ & = 0.9 - 0.75^2 \\ Var\begin{pmatrix}X \end{pmatrix} & = 0.3375 \end{aligned}\]

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3. We now calculate the standard deviation using the definition: \[\sigma = \sqrt{Var\begin{pmatrix} X\end{pmatrix}}\] Since \(Var\begin{pmatrix}X \end{pmatrix} = 0.3375\) we find: \[\sigma = \sqrt{0.3375} = 0.581\] The standard deviation is \(\sigma = 0.581\).
   This tells that, on average amongst all the different cell phones, we can expect the time taken for the app to launch to be \(0.581\) seconds more, or less, than the mean value \(\mu = 0.75\) seconds.