Find the value of the constant k so that the given function in a probability density function for a random variable over the
specified interval.
-$f ( x ) = 6 e ^ { - 5 x }$ over $[ 0 , k ]$
A) $\frac { 1 } { 5 } \ln \frac { 5 } { 6 }$
B) $- \frac { 1 } { 5 } \ln \frac { 11 } { 6 }$
C) $\frac { 1 } { 5 } \ln 6$
D) $e ^ { - 5 x } \ln 6$

Question

Quizplus · Accepted Answer

For a function to be a probability density function (pdf) over an interval, the integral of the function over that interval must equal 1. The integral of $f(x) = 6e^{-5x}$ from 0 to k is calculated as follows:$$\int_{0}^{k} 6e^{-5x} dx = \left[ -\frac{6}{5}e^{-5x} \right]_{0}^{k} = -\frac{6}{5}e^{-5k} + \frac{6}{5}$$Setting this equal to 1 and solving for k:$$-\frac{6}{5}e^{-5k} + \frac{6}{5} = 1$$$$-\frac{6}{5}e^{-5k} = \frac{1}{5}$$$$e^{-5k} = -\frac{1}{6}$$This equation does not make sense as written because the exponential function cannot give a negative result. The correct steps should be:$$-\frac{6}{5}e^{-5k} + \frac{6}{5} = 1$$$$e^{-5k} = \frac{1}{6}$$Taking the natural logarithm of both sides:$$-5k = \ln\left(\frac{1}{6}\right)$$$$k = -\frac{1}{5}\ln\left(\frac{1}{6}\right) = \frac{1}{5}\ln(6)$$Therefore, the correct answer is $\frac{1}{5}\ln(6)$, which corresponds to choice C.

Find the Value of the Constant K So That the Given