1. **State the problem:** We have a probability density function (PDF) defined as $$f(x) = \begin{cases} 6.22 e^{-x^{6.22}} x^{5.22}, & x \geq 0 \\ 0, & x < 0 \end{cases}$$ We want to understand the shape of this PDF and the relationship between the median and mode given by $$m = \left(1 - \frac{1}{6.22}\right)^{\frac{1}{6.22}}$$ 2. **Formula and explanation:** This PDF resembles a Weibull distribution with shape parameter $k=6.22$ and scale parameter $\lambda=1$. The mode of a Weibull distribution is given by $$\text{mode} = \left(1 - \frac{1}{k}\right)^{\frac{1}{k}}$$ which matches the given formula. 3. **Intermediate work:** Calculate the mode value: $$1 - \frac{1}{6.22} = 1 - 0.16077 = 0.83923$$ $$\text{mode} = 0.83923^{\frac{1}{6.22}}$$ Using logarithms: $$\ln(\text{mode}) = \frac{1}{6.22} \ln(0.83923) = \frac{1}{6.22} \times (-0.175) = -0.0281$$ $$\text{mode} = e^{-0.0281} \approx 0.9723$$ 4. **Interpretation:** The mode is approximately $0.9723$, meaning the PDF peaks near this value. 5. **Summary:** The PDF is zero for $x<0$ and for $x \geq 0$ it rises to a peak near $x=0.9723$ and then decays exponentially. The formula given correctly computes the mode of this distribution. **Final answer:** $$\boxed{\text{mode} \approx 0.9723}$$