Answer to Question #347457 in Statistics and Probability for Moro

Question

QUESTION 26

A local fire station receives on average 8.5 emergency telephone calls per hour. Assume that

these calls are Poisson distributed. Calculate the probability that

(a) the fire station will get nine calls during one hour. (2)

(b) the fire station will get five to seven (inclusive) calls during one hour. (3)

(c) the fire station will get at least 4 calls during one hour. (3)

(d) the fire station will get more than 6 calls during one hour.

Answer 1

(a)

$$P(X=9)=\dfrac{e^{-8.5}(8.5)^9}{9!}=0.129869$$

(b)

$$P(5\le X\le 7)=P(X=5)+P(X=6)$$

$$+P(X=7)=\dfrac{e^{-8.5}(8.5)^5}{5!}+\dfrac{e^{-8.5}(8.5)^6}{6!}$$

$$+\dfrac{e^{-8.5}(8.5)^7}{7!}=0.311233$$

(c)

$$P(X\ge 4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)=1-\dfrac{e^{-8.5}(8.5)^0}{0!}$$

$$-\dfrac{e^{-8.5}(8.5)^1}{1!}-\dfrac{e^{-8.5}(8.5)^2}{2!}$$

$$-\dfrac{e^{-8.5}(8.5)^3}{3!}=0.969891$$

(d)

$$P(X>6)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)-P(X=4)$$

$$-P(X=5)-P(X=6)=1-\dfrac{e^{-8.5}(8.5)^0}{0!}$$

$$-\dfrac{e^{-8.5}(8.5)^1}{1!}-\dfrac{e^{-8.5}(8.5)^2}{2!}$$

$$-\dfrac{e^{-8.5}(8.5)^3}{3!}-\dfrac{e^{-8.5}(8.5)^4}{4!}$$

$$-\dfrac{e^{-8.5}(8.5)^5}{5!}-\dfrac{e^{-8.5}(8.5)^6}{6!}=0.743822$$