Question

question 3
1 pts
suppose that on average, 2.4 patients come into the emergency room of a hospital with a serious injury during a 3 - hour shift. what is the probability of having exactly one patient come into the hospital during a 3 - hour shift?
hint: if you decide that you need values of ( e^{-x} )...use one of the following:
( e^{-9}=0.0001 ) ( e^{-0.1}=0.9048 ) ( e^{-1.2}=0.3012 ) ( e^{-3.2}=0.0407 ) ( e^{-2.3}=0.5134 )
( e^{-8}=0.0003 ) ( e^{-0.2}=0.8187 ) ( e^{-1.5}=0.2231 ) ( e^{-3.6}=0.0273 ) ( e^{-3.4}=0.4724 )
( e^{-6}=0.0025 ) ( e^{-0.3}=0.7408 ) ( e^{-1.6}=0.2019 ) ( e^{-4.8}=0.0082 ) ( e^{-5.6}=0.4346 )
( e^{-5}=0.0067 ) ( e^{-0.4}=0.6703 ) ( e^{-1.8}=0.1653 ) ( e^{-1.9}=0.7165 ) ( e^{-3.8}=0.6873 )
( e^{-4}=0.0183 ) ( e^{-0.5}=0.6065 ) ( e^{-2.1}=0.1225 ) ( e^{-1.4}=0.7788 ) ( e^{-5.8}=0.5353 )
( e^{-3}=0.0498 ) ( e^{-0.6}=0.5488 ) ( e^{-2.4}=0.0907 ) ( e^{-1.6}=0.8465 ) ( e^{-7.8}=0.4169 )
( e^{-2}=0.1353 ) ( e^{-0.8}=0.4493 ) ( e^{-2.5}=0.0821 ) ( e^{-1.8}=0.8825 ) ( e^{-4.3}=0.2636 )
( e^{-1}=0.3679 ) ( e^{-0.9}=0.4066 ) ( e^{-2.7}=0.0672 ) ( e^{-1.9}=0.8948 ) ( e^{-5.3}=0.1889 )
( 0.1195 )
( 0.0907 )
( 0.3595 )
( 0.0498 )
( 0.2177 )
( 0.1494 )

Explanation:

Step1: Identify the distribution

This is a Poisson distribution problem. The Poisson probability formula is $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the average number of events, and $k$ is the number of events we are interested in. Here, $\lambda = 2.4$ (average number of patients in a 3 - hour shift) and $k = 1$ (exactly one patient).

Step2: Substitute values into the formula

First, calculate $e^{-\lambda}$. Since $\lambda = 2.4$, we need to find $e^{-2.4}$. From the given hints, $e^{-2.4}=0.0907$. Then, $\lambda^k=(2.4)^1 = 2.4$, and $k! = 1! = 1$. Now substitute these values into the formula: $P(X = 1)=\frac{(2.4)^1\times e^{-2.4}}{1!}=\frac{2.4\times0.0907}{1}$.

Step3: Calculate the result

Multiply 2.4 and 0.0907: $2.4\times0.0907 = 0.21768\approx0.2177$.

Answer:

0.2177