HW7

1

Xi is uniform from [1,K] so total number of combination for x1 …xn is k^n Y is min{x1,x2…xn} if Y = 1 then there are (k-1)^n ways if Y =2 then there are (k-1)^n- (k-2)^n ways if Y = 3 then there are (k-2)^n - (k-3)^n ways

SO P(Y=y) = [(k-y+1)^n -(k-y)^n]/K^n

2

this is a geometric distribuction since it’s number of success until fill P(x=k) = p^(1-k)p E(x) = 1/p = 10 Var(x) = 1-p/p^2 p = 0.1 P(x>8) = (1-0.1)^8 = 0.4304672 Var(x) = 0.9/0.01 = 90 SD = 9.486833

pgeom(7,0.1,lower.tail = FALSE)

## [1] 0.4304672

sqrt(0.9/0.01)

## [1] 9.486833

3

P(x=k) = ae^(-ak) E(x) = 1/a Var(x) = 1/a^2 = 1/0.01 = 10 P(x>8) = 1-(1-e^-.08) = 0.449329

pexp(8,0.1,lower.tail = FALSE)

## [1] 0.449329

sqrt(1/0.01)

## [1] 10

3

p(x=k) = C(n,k)p^k(1-p)(n-k) E(x)= np =0.8 Var(x) = npq

pbinom(0,8,0.1)

## [1] 0.4304672

E = 0.8
sqrt(8*0.1*0.9)

## [1] 0.8485281

4

P(x=k) = e^(-a)ak/K! E(x) = a = 0.1X8 = 0.8 Var(x) = a = 0.8

ppois(0,0.8)

## [1] 0.449329

sqrt(0.8)

## [1] 0.8944272