Using (6.57), we obtain

\text{prob} \left(v \ \geq \ V\right) \ = \ \exp \ \left[- \frac{\pi }{4} \left(\frac{V}{\bar{v}}\right)^{2} \right] \quad \left(\text{Rayleigh}\right) (6.57)

\begin{matrix} \text{prob}\left(v \ \geq \ 6.5\right) \ = \ \exp \ \left[- \frac{\pi }{4} \ \left(\frac{6.5}{8} \right)^{2}\right] \ = \ 0.59542 \\ \text{prob}\left(v \ \geq \ 7.5\right) \ = \ \exp \ \left[- \frac{\pi }{4} \ \left(\frac{7.5}{8} \right)^{2}\right] \ = \ 0.50143 \end{matrix}

So, the probability that the wind is between 6.5 and 7.5 m/s is

\text{prob}\left(6.5 \ \leq \ v \ \leq \ 7.5\right) \ = \ 0.59542 \ − \ 0.50143 \ = \ 0.09400

From (6.45), we will approximate the probability that the wind is at 7 m/s to be

f\left(v\right) \ = \ \frac{\pi \ v}{2 \bar{v}^{2}} \ \exp \ \left[-\frac{\pi }{4} \ \left(\frac{v}{\bar{v}}\right)^{2} \right]

so,

f\left(7 \ {m}/{s}\right) \ = \ \frac{\pi \ \cdot \ 7}{2 \ \cdot \ 8^{2}} \ \exp \ \left[-\frac{\pi }{4} \ \left(\frac{7}{8}\right)^{2} \right] \ = \ 0.09416

The approximation 0.09416 is only 0.2% higher than the correct value of 0.09400.