Question

(3.4) This question is about a continuous probability dis- tribution known as the exponential distribution Let x be a continuous random variable that can take any value x 20. A quantity is said to be exponen- tially distributed if it takes values between r and r + dr with probability where A and A are constants. (a) Find the value of A that makes P() a well- defined continuous probability distribution so that Jo o P(x) dx = 1 (b) Show that the mean value of the probability distribution is(z)-/ xP(z) d. λ. (c) Find the variance and standard deviation of this probability distribution. Both the expo- nential distribution and the Poisson distribu- tion are used to describe similar processes but for the exponential distribution r is the actual time between, for example, successive radioactive decays, successive molecular col- lisions, or successive horse-kicking incidents rather than, as with the Poisson distribution, r being simply the number of such events in a specified interval)

Answer 1

The solution is given in the attached image.

Final answers are boxed.

P (x0 dx = A e^- x/lambda dx integral^infinity_0 p (x) dx = 1 rightdoublearrow A integral^infinity_0 e^- x/lambda dx = 1 (- lambda A) times e^- x/lambda integral_infinity_0 = 1 - lambda A times (- 1) = 1 A = 1/lambda < x > = 1/lambda integral^infinity_0 e^- x/lambda dx = 1/lambda [(- lambda) x e^- x/lambda integral^infinity_0 + integral^infinity_0 e^- x/lambda dx] 1 times lambda e^- x/lambda integral^infinity_0 (- lambda) = (- lambda) (- 1) = lambda sigma^2 = variance = < x^2 > - < x >^2 < x^2 > = 1/lambda integral^infinity_0 x^2 e^- x/lambda dx = 1/lambda [x^2 e^- x/lambda (- lambda) integral^infinity_0 + lambda integral^infinity_0 2 x e^- x/lambda dx] = 1/lambda times 2 lambda times lambda^2 = [2 lambda^2] rightadoublearrow variance = 2 lambda^2 - lambda = [lambda^2] S D = squareroot variance = squareroot lambda^2 = lambda