the indicated probabilities using the geometric​ distribution, the Poisson​ distribution, or the binomial distribution. Then determine if the events are unusual. If​ convenient, use the appropriate probability table or technology to find the probabilities. The mean number of births per minute in a country in a recent year was about seven. Find the probability that the number of births in any given minute is​ (a) exactly six​, ​(b) at least six​, and​ (c) more than six. ​(a) P(exactly six​)= enter your response here ​(Round to four decimal places as​ needed.)

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:

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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