Question

The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second. Use the central limit theorem to estimate the probability that more than 950 message arrive in one minute.

Answer 1

Answer:

0.048 is the probability that more than 950 message arrive in one minute.

Step-by-step explanation:

We are given the following information in the question:

The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.

Let X be the number of messages arriving at a multiplexer.

Mean = 15

For poison distribution,

Mean = Variance = 15

$\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{15} = 3.872$

From central limit theorem, we have:

$z = \displaystyle\frac{x-n\mu}{\sigma\sqrt{n}}$

where n is the sample size.

Here, n = 1 minute = 60 seconds

P(x > 950)

$P( x > 950) = P( z > \displaystyle\frac{950 - (60)(15)}{\sqrt{(15)(60)}}) = P(z > 1.667)$

$= 1 - P(z \leq 1.667)$

Calculation the value from standard normal z table, we have,  

$P(x > 950) = 1 - 0.952 = 0.048$

0.048 is the probability that more than 950 message arrive in one minute.