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3241004551 [841]
3 years ago
6

Suppose that the log-ons to a computer network follow a Poisson process with an average of 3 counts per minute.

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0

Answer:

a) 20 second

b) 20 second

c) 1 minute

Step-by-step explanation:

Let X be the exponential random variable arising from the Poisson process with the rate λ = 3 counts per minute. Its cdf is then given by:  

F(x) = I - e^-3x, x >= 0  

Calculate the mean and the standard deviation of the random variable X as follows:  

Е(X) =I/λ =I/3=20 second

std (x) =√1/λ^2=1/3=20 second

For the part c), write down the equation:

0.95 = P(X < x) = F(x) = I - e^-3x

and solve the equation for x to obtain:

x = -1/3 In 0.05 = 0.9985 ≅  1 minute

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An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
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Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

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Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

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However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

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