Definition of Beta Type I Distribution

A continuous random variable X is said to have a beta type I distribution with parameters α and β if its p.d.f. is given by f(x)={1B(α,β)xα1(1x)β1,0x1;0,Otherwise. where,

  • B(α,β)=ΓαΓβΓ(α+β)=10xα1(1x)β1dx is a beta function and

  • Γ(α) is a gamma function.

Mean of Beta Type I Distribution

The mean of beta type I distribution is E(X)=αα+β.

Variance of Beta Type I Distribution

The variance of beta type I distribution is V(X)=αβ(α+β)2(α+β+1).

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 α=5 and β=2.

Calculate

a. E(X) and V(X),

b. P(X0.2)

c. P(0.2X0.4)

Solution

Let X denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that Xβ1(α,β).

The probability density function of X is

f(x)=1B(5,2)x51(1x)21;0x1=Γ(5+2)Γ(5)Γ(2)x4(1x)1;0x1=6!4!1!x4(1x)1;0x1=30x4(1x);0x1

a. Mean and variance of X

E(X)=αα+β=55+2=0.7143.

V(X)=αβ(α+β)2(α+β+1)=5×2(5+2)2(5+2+1)=0.0255

b. P(X0.2)

P(X0.2)=0.20f(x)dx=300.20x4(1x)dx=300.20(x4x5)dx=30[x55x66]0.20=30[0.2550.266]=0.0016

c. P(0.2X0.4)

P(0.2X0.4)=0.40.2f(x)dx=300.40.2x4(1x)dx=300.40.2(x4x5)dx=30[x55x66]0.40.2=30[0.4550.4660.255+0.266]=0.0394

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 α=3,β=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 Xβ1(α,β).

The probability density function of X is

f(x)=1B(3,2)x31(1x)21;0x1=Γ(3+2)Γ(3)Γ(2)x2(1x)1;0x1=4!2!1!x2(1x)1;0x1=12x2(1x);0x1

a. Mean of X is

E(X)=αα+β=33+2=0.6. Variance of X is V(X)=αβ(α+β)2(α+β+1)=3×2(3+2)2(3+2+1)=0.04. Thus the standard deviation of X is σ=V(X)=0.04=0.2.

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(X0.8)

P(X0.8)=10.8f(x)dx=1210.8x2(1x)dx=1210.8(x2x3)dx=12[x33x44]10.8=12[1331440.833+0.844]=0.1808

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(0.6X0.8)

P(0.6X0.8)=0.80.6f(x)dx=120.80.6x2(1x)dx=120.80.6(x2x3)dx=12[x33x44]0.80.6=12[0.8330.8440.633+0.644]=0.344

I hope this article helps you understand how to solve the numerical problems on Beta type I distribution.

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