Question

historical data indicates that rickenbacker airlines receives an average of 2.3 complaints per day.
what is the probability, rounded to four decimal places, that on a given day, rickenbacker airlines will receive less than 6 complaints?

Explanation:

Step1: Identify the distribution

This is a Poisson distribution problem. The formula for the Poisson probability mass function is \(P(X = k)=\frac{\lambda^{k}e^{-\lambda}}{k!}\), where \(\lambda\) is the average number of events per interval, and \(k\) is the actual number of events. We want to find \(P(X < 6)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)\) with \(\lambda = 2.3\).

Step2: Calculate each term

• For \(k = 0\):

\(P(X = 0)=\frac{2.3^{0}e^{-2.3}}{0!}=e^{-2.3}\approx0.1003\)

• For \(k = 1\):

\(P(X = 1)=\frac{2.3^{1}e^{-2.3}}{1!}=2.3e^{-2.3}\approx2.3\times0.1003 = 0.2307\)

• For \(k = 2\):

\(P(X = 2)=\frac{2.3^{2}e^{-2.3}}{2!}=\frac{5.29e^{-2.3}}{2}\approx\frac{5.29\times0.1003}{2}= 0.2653\)

• For \(k = 3\):

\(P(X = 3)=\frac{2.3^{3}e^{-2.3}}{3!}=\frac{12.167e^{-2.3}}{6}\approx\frac{12.167\times0.1003}{6}=0.2042\)

• For \(k = 4\):

\(P(X = 4)=\frac{2.3^{4}e^{-2.3}}{4!}=\frac{27.9841e^{-2.3}}{24}\approx\frac{27.9841\times0.1003}{24}=0.1172\)

• For \(k = 5\):

\(P(X = 5)=\frac{2.3^{5}e^{-2.3}}{5!}=\frac{64.36343e^{-2.3}}{120}\approx\frac{64.36343\times0.1003}{120}=0.0540\)

Step3: Sum up the probabilities

\(P(X < 6)=0.1003 + 0.2307+0.2653 + 0.2042+0.1172+0.0540=0.9717\)

Answer:

\(0.9717\)