Q. 6.34 - Elementary Statistics

Question

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?

Step-by-Step Solution

Answer

Part a. 

Part b. \(14.65%\)

Part a. Step 1. Given information

Random variable \(‘x’\) 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 \(0.25\))

                                                           \(=\)\((1-e^{-0.69\times0.5}-(1-e^{-0.69\times0.25}\)

                                                           \(=\)\((1-0.0317)-(1-0.1782)\)

                                                           \(=0.9683-0.8218\)

                                                           \(=0.1465\)

                                                           \(=14.65%\)