Question

Errors in measuring the time of arrival of a wave front from an acoustic source sometimes have an approximate beta distribution. Suppose that these errors, measured in microseconds, have approximately a beta distribution with α = 1 and β = 3.

What is the probability that the measurement error in a randomly selected instance is less than 0.8 µs?

Give the mean and standard deviation of the measurement errors in microseconds.

Answer 1

a)

f(x)=(1/B(α,β))*xα-1*(1-x)β-1 =

3*x^0*(1-x)^2

=3(1-x)2

P(X<0.8)= f(x )dx = 3(1-x)² dx =-3*(1-x)³/3|^0.8₀ =0.9920

b)

mean μ=E(X)=α/(α+β)=1/(1+3) =

0.2500

variance=σ2=	αβ/((α+β)2*(α+β+1))=	0.0375
standard deviation=σ=√σ2=		0.1936