Question

CAN SOMEONE PLEASE HELP ME WITH PART C BY SHOWING ALL WORK AND NOT USING SOFTWARE TO SOLVE THE GAMMA DISTRIBUTION FORMULA FOR PART C! I TRIED DOING IT ON MY OWN AND END UP WITH A PROBABILITY OF 3.51 x 10^-10 WHICH I'M NOT SURE IS RIGHT. THANK YOU

Audrey, an astronomer is searching for extra - solar planets using the technique of relativistic lensing. Though there are believed to be a very large number of planets that can be found this way, actually finding one takes time and luck; and finding one planet does not help at all with finding planets of other stars in the same part of the sky. Audrey is good at it, and finds one planet at a time, on average once every three months.

a.)

Find the expected value and standard deviation of the number of planets she will find in the next two years.

b.) When she finds her sixth new planet, she will be eligible for a prize. Find the expected value of the amount of time until she is eligible for that prize.

c.) Find the probability that she will become eligible for that prize within one year.

Answer 1

a)

Assuming the time T to find the next planet follows exponential distribution. T ~ Exp($\lambda$ = 1/3)

Average number of planet found in two years = (1/3) * 24 = 8 planets

The inter-arrival time between Poisson processes follows exponential distribution. Thus, number of planets she will find in the next two years will follow Poisson distribution with mean $\lambda$ = 8.

Expected value = $\lambda$ = 8

Standard deviation of the number of planets = = 2.828

V =V8

b)

The time until sixth new planet T6 will follow Gamma distribution with k = 6 and $\lambda$ = 1/3

Expected value of the amount of time until she is eligible for that prize = k / $\lambda$ = 6 / (1/3) = 18 months

c)

Probability that she will become eligible for that prize within one year = P(T6 < 12 months)

= P(T6 < 12)

CDF of Gamma distribution is,

PT6 < =) = rm 4* *** dt = Tas (ki, dr

where

is lower incomplete gamma function

(k, Ar)

and

71s +1,2) = sy(s, ) – re-

1 712/3 PT. 1 P(T6 <12) = T(6) lo *** 76-1e+ dt = 16 (6,12 * (1/3) = 576,4)

Now,

76,4) = 57(5,4) – 4e-4

7(5, 4) = 4,(4,4) – 44e-4

7(4,4) = 37(3,4) – 43e-4

7(3, 4) = 27(2,4) – 4e-4

7(2, 4) = 17(1,4) – 41-4

|(1,4) = 17-12-dt = + dt = 1--= 0.9816844

(2, 4) = 0.9816844 – 4e-4 = 0.9084218

7(3, 4) = 2 * 0.9084218 – 16e-4 = 1.523793

7(4,4) = 3 * 1.523793 – 64e-4 = 3.399178

7(5, 4) = 4 * 3.399178 – 256 *e-4 = 8.907908

7(6,4) = 5 * 8.907908 – 1024e-4 = 25.78433

P(T6 < 12) = 5(6,4) = 25.78433/120 = 0.2148694