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Sagot :
Using the Poisson distribution, it is found that the probabilities are:
a) 0.149
b) 0.6993
c) 0.5503
We are given only the mean, hence, the Poisson distribution is used.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
The mean is of 7 births per minute, hence [tex]\mu = 7[/tex].
Item a:
The probability is P(X = 6), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 6) = \frac{e^{-7}(7)^{6}}{(6)!} = 0.149[/tex]
Item b:
This probability is:
[tex]P(X \geq 6) = 1 - P(X < 6)[/tex]
In which:
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
Following the logic used in item a to find each probability, we have that [tex]P(X < 6) = 0.3007[/tex], hence:
[tex]P(X \geq 6) = 1 - P(X < 6) = 1 - 0.3007 = 0.6993[/tex]
Item c:
This probability is:
[tex]P(X > 6) = 1 - P(X \leq 6)[/tex]
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
We have that [tex]P(X \leq 6) = 0.4497[/tex], hence:
[tex]P(X > 6) = 1 - P(X \leq 6) = 1 - 0.4497 = 0.5503[/tex]
A similar problem is given at https://brainly.com/question/16912674
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