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 a approximately a beta distribution with α = 1 and β = 2



What is the probability that the measurement error in a randomly selected instance us less than 0.6 µs?



Give the mean and standard deviation of the measurement errors.

Answer 1

Answer:

a) P=0.84

b) Mean=0.33

Standard deviation=0.356

Step-by-step explanation:

The probabilty that the measurement error in a randomly selected instance us less than 0.6 µs is P=0.84.

The mean of a Beta(α = 1, β = 2) is

$\mu=\frac{\alpha}{\alpha+\beta}=\frac{1}{1+2}=0.33$

The standard deviation of a Beta(α = 1, β = 2) is

$\sigma=\sqrt{\frac{\alpha\beta}{(\alpha+\beta)^2*(\alpha+\beta+1)}}\\\\\\\sigma= \sqrt{\frac{1*2}{(1+2)^2*(1+2+1)}}=\sqrt{\frac{2}{(2)^2*(4)}}=\sqrt{\frac{2}{16} } =\sqrt{0.125}= 0.356$