Important Distributions (Chapter 5 - Section 5.1) - Question 14

Week07 Discussion
Solution (a):

We could appoximate this situation with a Poisson distribution.

lambda = 10000*(1/1000)
k = 0
e = exp(1)
  
p = (lambda^k/factorial(k)) * (e^(-1*lambda))
print (p)
## [1] 4.539993e-05

The expected value of \(S_{10000}\) is \(\lambda = 10000.\frac{1}{1000} = 10\)

\(\lambda = 10\)
\(k = 0\)

\(P(X = 0) \approx \frac{\lambda^k}{k!}.e^{-\lambda} \approx \frac{10^0}{0!}.e^{-10}\)
\(\approx e^{-10}\)

Solution (b):

The probability of finding at least one person is one minus the probability of finding zero people.

\(\lambda = \frac{N}{1000}\)
\(k = 0\)

\(\frac{1}{2} =\frac{\lambda^k}{k!}.e^{-\lambda} = \frac{{(\frac{N}{1000}})^0}{0!}.e^{-\frac{N}{1000}}\)

\(\frac{1}{2} = e^{-\frac{N}{1000}}\)

\(log (\frac{1}{2}) = -\frac{N}{1000}.log(e)\)

\(N\) = \(693.1471\)
\(N \approx 694\)