Resources for Continuous Probability Distributions
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Questions
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Video Tutorials
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HSC Questions
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Continuous Probability Distributions Theory
![The area under a probability density function is 1. \\ The mean \(\mu=\displaystyle\int_a^b x f(x) \,d x\)\\ The variance \(\sigma ^2=\displaystyle\int_a^b x^2f(x)\,dx - \mu^2\)\\ The standard deviation \(= \sigma\)\\ The cumulative distribution function (CDF) is given by \(F(x)=\displaystyle \int_a^x f(x) d x\) where \(y=f(x)\) is a PDF defined in the domain \([a, b]\).\\ The median is determined by letting \(F(x)=0.5\). \begin{multicols}{2} \textbf{Example 1}\\ For the continuous random variable with probability density function (PDF):\\ \(f(x)=\left\{\begin{array}{l} \dfrac{3}{7}\left(x^2+2\right), 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right.\)\\ find correct to four decimal places \(P(x \leqslant 0.6)\)\\ \textbf{Example 1 solution} \begin{align*} P(x \leq 0.6) \quad I & =\frac{3}{7} \int_0^{0.6} x^2+2\,d x \\ & =\frac{3}{7}\left[\frac{x^3}{3}+2 x\right]_0^{0.6} \\ & =\frac{3}{7}\left[\frac{(0.6)^3}{3}+2 \times 0.6-0\right] \\ & =0.5451 \end{align*} \textbf{Example 2}\\ For the continuous random variable with probability density function\\ \( f(x)=\left\{\begin{array}{cl} \dfrac{3}{7}\left(x^2+2\right), & 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right. \)\\ find correct to two decimal places\\ (i) the mean\\ (ii) the standard deviation.\\ \vfill \columnbreak \textbf{Example 2 solution} \level \begin{align*} \text { (i) } \quad \mu & =\int_0^1 x f(x) d x \\ & =\frac{3}{7} \int_0^1 x\left(x^2+2\right) d x \\ & =\frac{3}{7} \int_0^1 x^3+2 x \,d x \\ & =\frac{3}{7}\left[\frac{x^4}{4}+x^2\right]_0^1 \\ & =\frac{3}{7}\left[\frac{1}{4}+1-0\right]=\dfrac{15}{28} \end{align*} \begin{align*} \text{ii) } \quad \sigma^2 & =\int_0^1 x^2 f(x) d x -\mu^2 \\ & =\frac{3}{7} \int_0^1 x^2\left(x^2+2\right) d x -\left(\frac{15}{28}\right)^2 \\ & =\frac{3}{7} \int_0^1 x^4+2 x^2 \,d x -\left(\frac{15}{28}\right)^2\\ & =\frac{3}{7}\left[\frac{x^5}{5}+\frac{2 x^3}{3}\right]_0^1 -\left(\frac{15}{28}\right)^2\\ & =\frac{3}{7}\left[\frac{1}{5}+\frac{2}{3}\right] -\left(\frac{15}{28}\right)^2 \\ & =\frac{331}{3920} \\ \sigma & =\sqrt{\frac{331}{3920}}=0.29 \end{align*} \end{multicols} \newpage \begin{multicols}{2} \textbf{Example 3}\\ For the continuous random variable with probability density function \[ f(x)=\left\{\begin{array}{l} \frac{3}{4} x(2-x) ; 0 \leq x \leq 2 \\ 0, \text { otherwise } \end{array}\right. \] Find the mode.\\ \textbf{Example 3 solution}\\ $\begin{aligned} f(x) & =\frac{3}{4} x(2-x) \\ & =\frac{3}{2} x-\frac{3}{4} x^2 \\ f^{\prime}(x) & =\frac{3}{2}-\frac{3}{2} x \\ \text { Let } f^{\prime}(x)=0 \qquad & \frac{3}{2}-\frac{3}{2} x=0 \\ f^{\prime \prime}(x) & =-\frac{3}{2} \end{aligned}$\\ $\begin{aligned} & \therefore x=1 \text { will give a maximum value } \\ & \therefore x=1 \text { is the mode } \end{aligned}$\\ \vfill \columnbreak \textbf{Example 4}\\ For the continuous probability defined as \(f(x)=\dfrac{x^2}{21}\) on the interval \(1 \leq x \leq 4\), find the median (correct to two decimal places).\\ \textbf{Example 4 solution}\\ $\begin{aligned} C D F & =\displaystyle \int_1^x \frac{x^2}{21} d x \\ & =\frac{1}{21}\left[\frac{x^3}{3}\right]_1^x \\ & =\frac{1}{21}\left[\frac{x^3}{3}-\frac{1}{3}\right] \\ & =\frac{1}{63}\left(x^3-1\right)\\ \text{Let CDF}&=\frac{1}{2} \text{ to determine the median}\\ &\frac{1}{63}\left(x^3-1\right) =\frac{1}{2} \\ &x^3-1 =31.5 \\ &x =\sqrt[3]{32.5} \\ &x =3.19 \end{aligned}$\\ \end{multicols}](/media/r1qka3vb/26848.png)