Answer to Question #234779 in Statistics and Probability for Booblex

Question

b) 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?

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

[2 marks] [2 marks]

c) Suppose the earnings of a laborer, denoted by X, are given by the following probability distribution.

 P(xi) 0 8/27

xi

 1 12/27

2 6/27 3 1/27

Find the laborer’s expected earnings and the variance of his earnings.

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)

$$\lambda t=6$$

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

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

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

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

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

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

c)

$$\begin{matrix} x_i & 0 & 1 & 2 & 3\\ p(x_i) & 8/27 & 12/27 & 6/27 & 1/27 \end{matrix}$$

Check

$$\dfrac{8}{27}+\dfrac{12}{27}+\dfrac{6}{27}+\dfrac{1}{27}=1$$

$$E[X]=\dfrac{8}{27}(0)+\dfrac{12}{27}(1)+\dfrac{6}{27}(2)+\dfrac{1}{27}(3)=1$$

$$E[X^2]=\dfrac{8}{27}(0)^2+\dfrac{12}{27}(1)^2+\dfrac{6}{27}(2)^2+\dfrac{1}{27}(3)^2=\dfrac{5}{3}$$

$$Var(X)=E[X^2]-(E[X])^2=\dfrac{5}{3}-(1)^2=\dfrac{2}{3}$$