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Nina [5.8K]
3 years ago
9

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

t theorem to estimate the probability that more than 950 message arrive in one minute.
Mathematics
1 answer:
Mila [183]3 years ago
3 0

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.

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The interval notation for the domain is [\frac{23}{3},\infty  ].

Step-by-step explanation:

Consider the provided information.

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We need to find the value of f\left(g\left(x\right)\right).

Put the value of g(x) in  f\left(g\left(x\right)\right).

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Here, the value of 3x-23 should be greater or equal to 0 as the square root of a negative number is not real.

Domain= 3x-23\geq0

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The value of x is all real number greater than \frac{23}{3}.

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