Question

In: Statistics and Probability

1. Suppose the number of messages arriving on a communication line seems to be describable by...

1. Suppose the number of messages arriving on a communication line seems to be describable by a Poisson Random Variable with an average arrival rate of 10 messages/second. Determine the following

a. The probability that no messages arrive in a one second period.

b. The probability that 1 message arrives in a one second period.

c. The probability that 2 messages arrives in a one second period.

d. The probability that 3 messages arrives in a one second period.

e.The probability that 10 messages arrive in a one second period.

f. The probability that 4or more messages arrive in a one second period

Solutions

Expert Solution

(a)

Probability Density Function of Poisson Distribution with mean = $\lambda$ = 10 is given by:

$P(x)=\frac{e^{-10}\times 10^{x}}{x!}$ ,

for x = 0, 1, 2,..

For x =0:

$P(x=0)=\frac{e^{-10}\times 10^{0}}{0!}=\frac{0.00004540\times 1}{1}=0.00004540$

(b)

$P(x=1)=\frac{e^{-10}\times 10^{1}}{1!}=\frac{0.00004540\times 10}{1}=0.0004540$

(c)

$P(x=2)=\frac{e^{-10}\times 10^{2}}{2!}=\frac{0.00004540\times 100}{2}=0.00227$

(d)

$P(x=3)=\frac{e^{-10}\times 10^{3}}{3!}=\frac{0.00004540\times 1000}{6}=0.00757$

(e)

$P(x=10)=\frac{e^{-10}\times 10^{10}}{10!}=\frac{0.00004540\times 10^{10}}{3628800}=0.12511$

(f)

P(X $\geq$ 4) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

$P(x=0)=\frac{e^{-10}\times 10^{0}}{0!}=\frac{0.00004540\times 1}{1}=0.00004540$

$P(x=1)=\frac{e^{-10}\times 10^{1}}{1!}=\frac{0.00004540\times 10}{1}=0.0004540$

$P(x=2)=\frac{e^{-10}\times 10^{2}}{2!}=\frac{0.00004540\times 100}{2}=0.00227$

$P(x=3)=\frac{e^{-10}\times 10^{3}}{3!}=\frac{0.00004540\times 1000}{6}=0.00757$

So,

P(X $\geq$ 4) = 1 - 0.01034 = 0.98966

So,

Answer is:

0.98966

orchestra answered 2 years ago

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