Beta Type I Distribution Examples
Beta Type I Distribution Calculator
Use this calculator to find the probability density and cumulative probabilities for Beta Type I distribution with parameter $\alpha$ and $\beta$.
Beta Type I Distribution Calculator  

First Parameter $\alpha$:  
Second Parameter $\beta$  
Value of x  
Results  
Probability density : f(x)  
Probability X less than x: P(X < x)  
Probability X greater than x: P(X > x)  
How to calculate probabilities of Beta Type I Distribution?
Step 1 – Enter the first parameter $\alpha$
Step 2 – Enter the second parameter $\beta$
Step 3 – Enter the value of $x$
Step 4 – Click on "Calculate" button to get beta type I distribution probabilities
Step 5 – Gives the output probability density at $x$ for beta type I distribution
Step 6 – Gives the output probability $X < x$ for beta type I distribution
Step 7 – Gives the output probability $X > x$ for beta type I distribution.
Definition of Beta Type I distribution
The probability density function of beta type I distribution with parameter $\alpha$ and $\beta$ is
$$ \begin{align*} f(x) &= \begin{cases} \frac{1}{B(\alpha,\beta)}x^{\alpha1}(1x)^{\beta1}, & 0\leq x\leq 1 \\ 0, & Otherwise. \end{cases} \end{align*} $$
where,

$B(\alpha,\beta) =\frac{\Gamma (\alpha) \Gamma (\beta)}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha1}(1x)^{\beta1}\; dx$ is a beta function and

$\Gamma \alpha$ is a gamma function.
Beta Type I Distribution Example 1
Suppose the proportion $X$ of surface area in a randomly selected quadrant that is covered by a certain plant has a beta distribution with $\alpha=5$ and $\beta = 2$.
Calculate
a. $E(X)$ and $V(X)$,
b. $P(X\leq 0.2)$
c. $P(0.2\leq X\leq 0.4)$
Solution
Let $X$ denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that $X\sim \beta_1(\alpha, \beta)$.
The probability density function of $X$ is
$$ \begin{aligned} f(x)&= \frac{1}{B(5,2)}x^{51}(1x)^{21}; 0\leq x\leq 1\\ &=\frac{\Gamma(5 + 2)}{\Gamma(5)\Gamma(2)}x^{4}(1x)^{1}; 0\leq x\leq 1\\ &=\frac{6!}{4!1!}x^{4}(1x)^{1};0\leq x\leq 1\\ &=30x^{4}(1x);0\leq x\leq 1 \end{aligned} $$
a. Mean and variance of $X$
The mean of $X$ is
$$ \begin{aligned} E(X) &=\dfrac{\alpha}{\alpha+\beta}\\ &=\dfrac{5}{5 + 2}\\ &=0.7143. \end{aligned} $$
The variance of $X$ is
$$ \begin{aligned} V(X) &=\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\\ &=\dfrac{5\times 2}{(5 + 2)^2(5 + 2+1)}\\ &=0.0255. \end{aligned} $$
b. The probability that the surface area in a randomly selected quadrant that is covered by a certain plant is less than 0.2 is $P(X\leq 0.2)$.
$$ \begin{aligned} P(X\leq 0.2)&= \int_0^{0.2}f(x)\; dx\\ &= 30\int_0^{0.2}x^{4}(1x)\; dx\\ &= 30\int_0^{0.2}(x^{4}x^{5})\; dx\\ &= 30\bigg[\frac{x^{5}}{5}\frac{x^{6}}{6}\bigg]_0^{0.2}\\ &= 30\bigg[\frac{0.2^{5}}{5}\frac{0.2^{6}}{6}\bigg]\\ &=0.0016 \end{aligned} $$
c. The probability that the surface area in a randomly selected quadrant that is covered by a certain plant is between 0.2 and 0.4 is $P(0.2\leq X\leq 0.4)$
$$ \begin{aligned} P(0.2\leq X\leq 0.4)&= \int_{0.2}^{0.4}f(x)\; dx\\ &= 30\int_{0.2}^{0.4}x^{4}(1x)\; dx\\ &= 30\int_{0.2}^{0.4}(x^{4}x^{5})\; dx\\ &= 30\bigg[\frac{x^{5}}{5}\frac{x^{6}}{6}\bigg]_{0.2}^{0.4}\\ &= 30\bigg[\frac{0.4^{5}}{5}\frac{0.4^{6}}{6}\frac{0.2^{5}}{5}+\frac{0.2^{6}}{6}\bigg]\\ &=0.0394 \end{aligned} $$
Beta Type I Distribution Example 2
The proportion of a brand of a television set that requires repair during the first year is a random variable having a beta distribution with $\alpha = 3$ ,$\beta = 2$.
a. Find the average and standard deviation of proportion of a brand of a television set that requires repair during the first year.
b. What is the probability that at least 80% of the new models sold this year will require repairs during the first year of operation?
c. What is the probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation?
Solution
Let $X$ denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that $X\sim \beta_1(\alpha, \beta)$.
The probability density function of $X$ is
$$ \begin{aligned} f(x)&= \frac{1}{B(3,2)}x^{31}(1x)^{21}; 0\leq x\leq 1\\ &=\frac{\Gamma(3 + 2)}{\Gamma(3)\Gamma(2)}x^{2}(1x)^{1}; 0\leq x\leq 1\\ &=\frac{4!}{2!1!}x^{2}(1x)^{1};0\leq x\leq 1\\ &=12x^{2}(1x);0\leq x\leq 1 \end{aligned} $$
a. Mean of $X$ is
$$ \begin{aligned} E(X) &=\dfrac{\alpha}{\alpha+\beta}\\ &=\dfrac{3}{3 + 2}\\ &=0.6. \end{aligned} $$
Variance of $X$ is
$$ \begin{aligned} V(X) &=\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\\ &=\dfrac{3\times 2}{(3 + 2)^2(3 + 2+1)}\\ &=0.04. \end{aligned} $$
Thus the standard deviation of $X$ is
$$ \begin{aligned} \sigma &= \sqrt{V(X)}\\ &=\sqrt{0.04}\\ &=0.2. \end{aligned} $$
b. The probability that at least 80% of the new models sold this year will require repairs during the first year of operation i.e., $P(X\geq 0.8)$
$$ \begin{aligned} P(X\geq 0.8)&= \int_{0.8}^{1}f(x)\; dx\\ &= 12\int_{0.8}^{1}x^{2}(1x)\; dx\\ &= 12\int_{0.8}^{1}(x^{2}x^{3})\; dx\\ &= 12\bigg[\frac{x^{3}}{3}\frac{x^{4}}{4}\bigg]_{0.8}^{1}\\ &= 12\bigg[\frac{1^{3}}{3}\frac{1^{4}}{4}\frac{0.8^{3}}{3}+\frac{0.8^{4}}{4}\bigg]\\ &=0.1808 \end{aligned} $$
c. The probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation i.e., $P(0.6\leq X\leq 0.8)$
$$ \begin{aligned} P(0.6\leq X\leq 0.8)&= \int_{0.6}^{0.8}f(x)\; dx\\ &= 12\int_{0.6}^{0.8}x^{2}(1x)\; dx\\ &= 12\int_{0.6}^{0.8}(x^{2}x^{3})\; dx\\ &= 12\bigg[\frac{x^{3}}{3}\frac{x^{4}}{4}\bigg]_{0.6}^{0.8}\\ &= 12\bigg[\frac{0.8^{3}}{3}\frac{0.8^{4}}{4}\frac{0.6^{3}}{3}+\frac{0.6^{4}}{4}\bigg]\\ &=0.344 \end{aligned} $$
Conclusion
In this tutorial, you learned about how to calculate probabilities of Beta Type I distribution. You also learned about how to solve numerical problems based on Beta Type I distribution.
To read more about the step by step tutorial on Beta Type I distribution refer the link Beta Type I Distribution. This tutorial will help you to understand Beta Type I distribution and you will learn how to derive mean, variance, harmonic mean and mode of Beta Type I distribution.
To learn more about other probability distributions, please refer to the following tutorial:
Let me know in the comments if you have any questions on Beta Type I Distribution Examples and your thought on this article.