Suppose that the log-ons to a computer network follow a Poisson process with an average of 3 counts per minute.
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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
The success probability = 5 % = 0.05
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
A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails
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
= - 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 = = 2.233
P (x≤100) = P = P(z<0.34) = 0.63307.