The uniformly distributed on the integers from 1 to k, the miniumum value of Y is one Xi from Xi’s:
Assume the distribution function P(Y=z) = min{X1,X2,X3….Xi}
Case 1: Assumption: Y=1, which is minimum sample Xi (Select one sample from 10 samples) x = {1,2,3,4,5} Size of all sample x = {1,2,3,4,5} : \(5^{10}\) Size of non-sample x = {1,2,3,4} : \((5-1)^{10}\) \(P(Y=1)= ((5-0)^{10}-(5-1)^{10} )/ 5 ^{10}\)
Case 2: Assumption: (Select 2 sample from 10 samples) x = {1,2,3,4,5} Size of all sample: \(5^{10}\) Size of non-sample = \((5^{10}-4^{10})+3^{10}\) \(P(Y=1)= (5^{10}-(5^{10}-4^{10})-3^{10})/ 5 ^{10}\)
Using avariables: \[k^{n} - (k^{n} - (k-1)^{n}) - (k-2)^{n} \] \[= (k-1)^{n}-(k-2)^{n} \]
When m=2; \[=((k+m-1)^{n} - (k-m)^{n})/k^{n} \]
Using geometric distribution: \(p(X=n) = (1-p)^{n-1}*p\) P(failure) = 1/10 =0.1 P(Success) = 0.9
\[P(X < 9)= 0.1+0.1*0.9+0.1*0.9^{2}+....+0.1*0.9^{7} =0.6125 \]
\[P(X\geqslant 9)= 1-0.6125 = 0.3874\]
1-pgeom(8,0.1)
## [1] 0.3874205
Exponential density:
\(\mu = 10\)
\(\lambda = \frac{1}{\mu}\)
\(\lambda = 0.1\)
\[P(T \leq 8) = 1-e^{-0.1*8} = 0.5506\]
\[P(T > 8) = 1-0.5506 = 0.4493\]
1-pexp(8,0.1)
## [1] 0.449329
\(\lambda = 0.1\)
Exp <- 1/0.1
Exp
## [1] 10
\(\lambda = 0.1\)
var <- 1/(0.1^2)
sd <- sqrt(var)
sd
## [1] 10
p=0.1 and q=0.9
\[P(X \leq 8) = 0.1^00.9^8 = 0.9^8 = 0.4304\]
pbinom(0,8,0.1)
## [1] 0.4304672
Expected value:
n=8
p=0.1
Exp1 <- n*p
Exp1
## [1] 0.8
Standard Deviation:
q=0.9
var1 <- n*p*q
sd1 <- sqrt(var1)
sd1
## [1] 0.8485281
n <- 8
p <- 1 / 10
#lambda
ld <- n * p
# success
x <- 0
p0 <- ((ld^x * ((exp(1))^(-1*ld)))/factorial(x))
p0
## [1] 0.449329
ppois(0, ld)
## [1] 0.449329
Expected value: \(E[X] =\lambda = 0.8\)
Standard Deviation:
var2 <- ld
sd2 <- sqrt(var2)
sd2
## [1] 0.8944272