# 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. 