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 and standard deviation 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.

The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]

a)

The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month

[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}

(24)\\\\=8[/tex]

(Here. t = 24)

For Poisson process mean and variance are same,

[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828

[/tex]

b)

For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter [tex]\lambda[/tex]. The sum of [tex]K[/tex] of these independent exponential has distribution Gamma [tex](K, \lambda)[/tex].

From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}

[/tex]

Calculate the expected value.

[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}

\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]

(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)

[tex]=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]

Hence, the required probability is 0.2149 or 21.49%