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Magazine Chapter 6: Q. 6.34 (page 261)
Desert Samaritan Hospital in Mesa, Arizona keeps record of its emergency-room traffic. Beginning at \(6:00\)PM on any given day, the elapsed time in hours until the first patient arrives is a variable with density curve \(y=6.9e^{-6.9x}\) for \(x>0\) and \(y=0\) otherwise. Here \(e\) is Euler's number which is approximately \(2.71828\). Most calculators have an \(e-\)key. Using calculus, it can be shown that the area under this density curve to the left of any number \(x\) greater than \(0\) equals \(1-e^{-6.9x}\).
a. Graph the density curve of this variable.
b. What percentage of the time does the first patient arrive between \(6:15\)PM and \(6:30\)PM?
Short Answer
Part a.
Part b. \(14.65%\)
Step by step solution
Part a. Step 1. Given information
Random variable \(鈥榵鈥橽) is the elapsed time in hours, until the first patient arrives on a given day beginning with \(6:00\) pm.
The density function is given as
\(
y=\left\{\begin{matrix}
6.9e^{-6.9x}& for x>0\\
0 & otherwise
\end{matrix}\right.\)
\(e=2.71828\)
Part a. Step 2. Calculation
The graph of the density function can be drawn as
Part b. Step 1. Calculation
The graph of the density function can be drawn as
The percentage of time does the first patient arrive between \(6:15\)pm and \(6:30\)pm will be
(Area between \(6:15\) and \(6:30\)) \(=\) (Area between \(0.25\) and \(0.5\))
\(=\) (Area to the left of \(0.5\)) - (Area to the left of \(0.25\))
\(=\)\((1-e^{-0.69\times0.5}-(1-e^{-0.69\times0.25}\)
\(=\)\((1-0.0317)-(1-0.1782)\)
\(=0.9683-0.8218\)
\(=0.1465\)
\(=14.65%\)
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