Answer to Question #234372 in Statistics and Probability for mushota

Question

Phone calls enter the ”support desk” of an electricity supplying company on the

average two every 3 minutes. If one assumes an approximate Poisson process:

(i) What is the probability of no calls in 3 minutes?

[2 marks]

(ii) What is the probability of utmost 6 calls in a 9 minute period?

Answer 1

Let $X=$ the number of calls: $X\sim Po(\lambda t).$

(i)

$$\lambda t=2$$

$$P(X=0)=\dfrac{e^{-\lambda t}(\lambda t)^0}{0!}=e^{-2}\approx0.135335$$

(ii)

$$\lambda t=6$$

$$P(X\leq6)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)+P(X=4)+P(X=5)+P(X=6)$$

$$=\dfrac{e^{-\lambda t}(\lambda t)^0}{0!}+\dfrac{e^{-\lambda t}(\lambda t)^1}{1!}+\dfrac{e^{-\lambda t}(\lambda t)^2}{2!}+\dfrac{e^{-\lambda t}(\lambda t)^3}{3!}$$

$$+\dfrac{e^{-\lambda t}(\lambda t)^4}{4!}+\dfrac{e^{-\lambda t}(\lambda t)^5}{5!}+\dfrac{e^{-\lambda t}(\lambda t)^6}{6!}$$

$$=e^{-6}(1+6+18+36+54+64.8+64.8)$$

$$=224.6e^{-6}\approx0.606303$$