16 Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the probability that the operator will miss at most one call if she takes a 5-minute coffee break.
From http://www.r-tutor.com/elementary-statistics/probability-distributions/poisson-distribution:
The Poisson distribution is the probability distribution of independent event occurrences in an interval. If \(\lambda\) is the mean occurrence per interval, then the probability of having \(x\) occurrences within a given interval is
\(f(x) = \frac{\lambda^xe^{-\lambda}}{x!}\)
In this case, \(\lambda = 0.01 * 300\) or \(\lambda = 3\) (the probability of a call multiplied by the interval, which is 5 minutes times 60 seconds/minute).
# Manual calculation
# Set e and lambda
e <- exp(1)
lda <- 0.01 * 300
# Chance of zero calls
c0 <- ((lda^0)*(e^-3)) / factorial(0)
# Chance of one call
c1 <- ((lda^1)*(e^-3)) / factorial(1)
# Add the two together
c0 + c1## [1] 0.1991483# Calculate using Base R ppois function
ppois(1, lambda=3)## [1] 0.1991483# Both methods equivalent
c0 + c1 == ppois(1, lambda=3)## [1] TRUE