Beta Type-I Distribution Calculator with 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

$$ \begin{align*} f(x) &= \begin{cases} \frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}, & 0\leq x\leq 1 \\ 0, & Otherwise. \end{cases} \end{align*} $$

where,

  • $B(\alpha,\beta) =\frac{\Gamma (\alpha) \Gamma (\beta)}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$ 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

$$ \begin{aligned} f(x)&= \frac{1}{B(5,2)}x^{5-1}(1-x)^{2-1}; 0\leq x\leq 1\\ &=\frac{\Gamma(5 + 2)}{\Gamma(5)\Gamma(2)}x^{4}(1-x)^{1}; 0\leq x\leq 1\\ &=\frac{6!}{4!1!}x^{4}(1-x)^{1};0\leq x\leq 1\\ &=30x^{4}(1-x);0\leq x\leq 1 \end{aligned} $$

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 $P(X\leq 0.2)$.

$$ \begin{aligned} P(X\leq 0.2)&= \int_0^{0.2}f(x)\; dx\\ &= 30\int_0^{0.2}x^{4}(1-x)\; dx\\ &= 30\int_0^{0.2}(x^{4}-x^{5})\; dx\\ &= 30\bigg[\frac{x^{5}}{5}-\frac{x^{6}}{6}\bigg]_0^{0.2}\\ &= 30\bigg[\frac{0.2^{5}}{5}-\frac{0.2^{6}}{6}\bigg]\\ &=0.0016 \end{aligned} $$

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 $P(0.2\leq X\leq 0.4)$

$$ \begin{aligned} P(0.2\leq X\leq 0.4)&= \int_{0.2}^{0.4}f(x)\; dx\\ &= 30\int_{0.2}^{0.4}x^{4}(1-x)\; dx\\ &= 30\int_{0.2}^{0.4}(x^{4}-x^{5})\; dx\\ &= 30\bigg[\frac{x^{5}}{5}-\frac{x^{6}}{6}\bigg]_{0.2}^{0.4}\\ &= 30\bigg[\frac{0.4^{5}}{5}-\frac{0.4^{6}}{6}-\frac{0.2^{5}}{5}+\frac{0.2^{6}}{6}\bigg]\\ &=0.0394 \end{aligned} $$

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

$$ \begin{aligned} f(x)&= \frac{1}{B(3,2)}x^{3-1}(1-x)^{2-1}; 0\leq x\leq 1\\ &=\frac{\Gamma(3 + 2)}{\Gamma(3)\Gamma(2)}x^{2}(1-x)^{1}; 0\leq x\leq 1\\ &=\frac{4!}{2!1!}x^{2}(1-x)^{1};0\leq x\leq 1\\ &=12x^{2}(1-x);0\leq x\leq 1 \end{aligned} $$

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., $P(X\geq 0.8)$

$$ \begin{aligned} P(X\geq 0.8)&= \int_{0.8}^{1}f(x)\; dx\\ &= 12\int_{0.8}^{1}x^{2}(1-x)\; dx\\ &= 12\int_{0.8}^{1}(x^{2}-x^{3})\; dx\\ &= 12\bigg[\frac{x^{3}}{3}-\frac{x^{4}}{4}\bigg]_{0.8}^{1}\\ &= 12\bigg[\frac{1^{3}}{3}-\frac{1^{4}}{4}-\frac{0.8^{3}}{3}+\frac{0.8^{4}}{4}\bigg]\\ &=0.1808 \end{aligned} $$

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., $P(0.6\leq X\leq 0.8)$

$$ \begin{aligned} P(0.6\leq X\leq 0.8)&= \int_{0.6}^{0.8}f(x)\; dx\\ &= 12\int_{0.6}^{0.8}x^{2}(1-x)\; dx\\ &= 12\int_{0.6}^{0.8}(x^{2}-x^{3})\; dx\\ &= 12\bigg[\frac{x^{3}}{3}-\frac{x^{4}}{4}\bigg]_{0.6}^{0.8}\\ &= 12\bigg[\frac{0.8^{3}}{3}-\frac{0.8^{4}}{4}-\frac{0.6^{3}}{3}+\frac{0.6^{4}}{4}\bigg]\\ &=0.344 \end{aligned} $$

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:

Probability distributions

Let me know in the comments if you have any questions on Beta Type I Distribution Examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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