Question

In a recent year​ (365 days), a hospital had 5705

Answer 1

Using the Poisson distribution, it is found that:

a) The mean is of 15.63.

b) The probability is of 0.0911 = 9.11%.

c) The probability is $e^{-15.63}$, which is less than 0.05, hence 0 births in a single day would be a significantly low number of​ births.

What is the Poisson distribution?

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

• x is the number of successes.

• e = 2.71828 is the Euler number.

• $\mu$ is the mean in the given interval.

Item a:

The mean is given by:

$\mu = \frac{5705}{365} = 15.63$

Item b:

The probability is P(X = 17), hence:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 17) = \frac{e^{-15.63}(15.63)^{17}}{(17)!} = 0.0911$

The probability is of 0.0911 = 9.11%.

Item c:

The probability is P(X = 0), hence:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-15.63}(15.63)^{0}}{(0)!} = e^{-15.63}$

The probability is $e^{-15.63}$, which is less than 0.05, hence 0 births in a single day would be a significantly low number of​ births.

More can be learned about the Poisson distribution at brainly.com/question/13971530

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Answer 2

Using the Poisson distribution, it is found that

a) The mean is 15.63.

b) The probability is$e^{-15.63}$ of 0.0911 = 9.11%.

c) The probability is, which is less than 0.05, hence 0 births in a single day would be a significantly low number of​ births.

What is the Poisson distribution?

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P\left(X=x\right)=\frac{e^{-\mu }\mu ^x}{x!}$

The parameters are

x is the number of successes.

e = 2.71828 is the Euler number.

$\mu$ is the mean in the given interval.

Item a:The mean is given by:

$\:\mu =\frac{5705}{365}$

Item b: The probability is P(X = 17), hence:

$P\left(X=17\right)=\frac{e^{-15.63\:}\left(15.63\right)\:^{17}}{17!}=0.0911$

The probability is of 0.0911 = 9.11%.

Item c:The probability is P(X = 0), hence:

$P\left(X=0\right)=\frac{e^{-15.63\:}\left(15.63\right)\:^{0}}{0!}=e^{-15.63}$

The probability is,$e^{-15.63}$ which is less than 0.05,

hence 0 births in a single day would be a significantly low number of​ births.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1