May 01'23
Exercise
The lifetime of a light bulb has density function, [math]f[/math], where [math]f(x)[/math] is proportional to
[[math]]
\frac{x^2}{1+x^3}, \, 0 \lt x \lt5, \, \textrm{and}, \, 0 \, \textrm{otherwise}.
[[/math]]
Calculate the mode of this distribution.
- 0.00
- 0.79
- 1.26
- 4.42
- 5.00
May 01'23
Solution: C
It is not necessary to determine the constant of proportionality. Let it be c. To determine the mode, set the derivative of the density function equal to zero and solve.
[[math]]
\begin{align*}
0 &= f^{'}(x) = \frac{d}{dx} cx^2(1+x^3)^{-1} = 2cx(1+x^3)^{-1} + cx^2[-(1+x^3)^{-2}]3x^2 \\
&= 2cx(1+x^3)-3cx^4 \\
&= 2cx + 2cx^4 -3cx^4 = 2cx - cx^4 \\
&= 2- x^3 \Rightarrow x = 2^{1/3} = 1.26.
\end{align*}
[[/math]]