Solution: For \(1 ≤ j ≤ k\), \(m(j) = \frac{(k−j+1)^n−(k−j)}{k^n}\)
In other words
As Y is the minimum value of \(X_i\)’s over all of the \(X_i\)’s then the distribution function would be \(m(j) = P(Y = j)\).
We will need to count the number of ways that we can assign the \(X_i\)’s to values between \(j\) and \(k\) with at least one \(X_i\) being assigned to \(j\) divided by the total number of possible ways to assign \(X_i\)’s to values between \(1\) and \(k\).
\(P(X=n) = q^{n-1}p\) where \(n = 8\) and \(p = 0.1\)
## [1] 0.04304672
\(E(X) = \frac{1}{p}\)
## [1] 10
\(D(X) = \frac{1-p}{p^2}\)
## [1] 9.486833
\(P(X \ge n) = e^{\frac{-k}{\mu}}\) where \(\lambda = 0.1\) and \(\mu = \frac{1}{\lambda} = 10\)
\(P(X \ge 8) = e^{\frac{-8}{10}}\)
## [1] 0.449329
\(E(X) = \frac{1}{\lambda}\)
## [1] 10
\(D(X)\)
## [1] 10
$b(n, p, k) =pkq{n-k} $
\(P(K = 0) = b(8, 0.1, 0)\)
## [1] 0.4304672
\(E(X) = np\)
## [1] 0.8
\(D(X) = \sqrt{npq}\)
## [1] 0.8485281
\(P(X = k) \approx \frac{\lambda k}{k!}e^{−\lambda}\) where lambda \(8 / 10 = 0.8\)
Failing after 8 years, that is 8 or more years – \(P(X \ge 8)\)
\(\lambda\) is the expected number of successes in 8 years which is 0
.
## [1] 0.550671
\(E(X) = \lambda = 0.8\)
\(D(X) = sqrt(\lambda)\)
## [1] 0.8944272
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