Answer to Question #93449 in Statistics and Probability for Tushar khohar

Question

Let the number of phone calls received by a switchboard during a 5-minute interval be a random variable X with probability function

f(x)=(e^-2)(2^x)/x! ; x=0,1,2,3...


a. Determine the probability that X equals 0, 1, 2, 3, 4, 5 and 6.
b. Determine the cumulative distribution function for these values of X.

Answer 1

a)$P(X=0)=\frac{2^0}{0!}e^{-2}=\frac{1}{1}e^{-2}=e^{-2}$

$P(X=1)=\frac{2^1}{1!}e^{-2}=\frac{2}{1}e^{-2}=2e^{-2}$

$P(X=2)=\frac{2^2}{2!}e^{-2}=\frac{4}{2}e^{-2}=2e^{-2}$

$P(X=3)=\frac{2^3}{3!}e^{-2}=\frac{8}{6}e^{-2}=\frac{4}{3}e^{-2}$

$P(X=4)=\frac{2^4}{4!}e^{-2}=\frac{16}{24}e^{-2}=\frac{2}{3}e^{-2}$

$P(X=5)=\frac{2^5}{5!}e^{-2}=\frac{32}{120}e^{-2}=\frac{4}{15}e^{-2}$

$P(X=6)=\frac{2^6}{6!}e^{-2}=\frac{64}{720}e^{-2}=\frac{4}{45}e^{-2}$

b)$P(X\le 0)=P(X=0)=e^{-2}$

$P(X\le 1)=P(X=0)+P(X=1)=3e^{-2}$

$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)=5e^{-2}$

$P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=\frac{19}{3}e^{-2}$

$P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=7e^{-2}$

$P(X\le 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=\frac{109}{15}e^{-2}$

$P(X\le 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=\frac{331}{45}e^{-2}$