Question-14

Suppose the number of visitors arriving at a zoo can be modeled to be Poisson distributed. On an average 20 visitors arrive per hour. Let \(X\) be the number of visitors arriving from 2pm to 4pm. Then the probability that at least 35 visitors will arrive in the given duration is

• \(\sum\limits_{k = 35}^{\infty}\dfrac{e^{-20}(20)^k}{k!}\)

• \(1-\sum\limits_{k = 0}^{34}\dfrac{e^{-20}(20)^k}{k!}\)

• \(\sum\limits_{k = 35}^{\infty}\dfrac{e^{-40}(40)^k}{k!}\)

• \(1-\sum\limits_{k = 0}^{34}\dfrac{e^{-40}(40)^k}{k!}\)

Hint

\(\lambda = 20\) per hour. Therefore, for 2pm - 4pm, \(\lambda\) will be updated as \(= 2\times20 = 40\) because there are 2 hours for 2pm - 4pm and use pmf of Poisson distribution.

Answer

• \(\sum\limits_{k = 35}^{\infty}\dfrac{e^{-20}(20)^k}{k!}\)

• \(1-\sum\limits_{k = 0}^{34}\dfrac{e^{-20}(20)^k}{k!}\)

• \(\sum\limits_{k = 35}^{\infty}\dfrac{e^{-40}(40)^k}{k!}\)

• \(1-\sum\limits_{k = 0}^{34}\dfrac{e^{-40}(40)^k}{k!}\)

Solution