Question

in a recent year (365 days), a hospital had 5693 births. a. find the mean number of births per day. b. find the probability that in a single day, there are 18 births. c. find the probability that in a single day, there are no births. would 0 births in a single day be a significantly low number of births? a. the mean number of births per day is 15.6 (round to one decimal place as needed.) b. the probability that, in a day, there are 18 births is (do not round until the final answer. then round to four decimal places as needed.)

Explanation:

Step1: Calculate the mean number of births per day

The total number of births in a year is 5693 and the number of days in a year is 365. The mean $\lambda$ is calculated by dividing the total number of births by the number of days. So, $\lambda=\frac{5693}{365}=15.6$.

Step2: Use the Poisson - distribution formula for probability

The Poisson - distribution formula is $P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}$, where $\lambda$ is the mean, $k$ is the actual number of occurrences, and $e\approx2.71828$.

For part b:

We want to find $P(X = 18)$ with $\lambda = 15.6$.
$P(X = 18)=\frac{e^{- 15.6}(15.6)^{18}}{18!}$
First, calculate $e^{-15.6}\approx2.71828^{-15.6}\approx1.03\times10^{-7}$.
$(15.6)^{18}$ is a large number. $(15.6)^{18}\approx1.09\times10^{21}$.
$18! = 18\times17\times\cdots\times1=6.40237\times10^{15}$.
$P(X = 18)=\frac{1.03\times10^{-7}\times1.09\times10^{21}}{6.40237\times10^{15}}\approx0.1749$.

For part c:

We want to find $P(X = 0)$ with $\lambda = 15.6$.
$P(X = 0)=\frac{e^{-15.6}(15.6)^{0}}{0!}$
Since $(15.6)^{0}=1$ and $0! = 1$, $P(X = 0)=e^{-15.6}\approx1.03\times10^{-7}$.
A value is significantly low if $P(X\leq k)\leq0.05$. Since $P(X = 0)=1.03\times10^{-7}\leq0.05$, 0 births in a single day is a significantly low number of births.

Answer:

a. 15.6
b. 0.1749
c. 1.03×10⁻⁷; Yes, 0 births in a single day is a significantly low number of births.