The expectance probability ( in percentage) for the peak value of 320 cumec from the following flood data and using Weibull formula is , will be _________
$\begin{matrix} Y e a r & 2001 & 2002 & 2003 & 2004 & 2005 F l o o d p e a k (i n m^{3} / s) & 430 & 450 & 320 & 310 & 290 \end{matrix}$
50

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Solution

The correct option is A 50
The given peak can be arranged in decreasing order as;
$\begin{matrix} Y e a r & 2001 & 2002 & 2003 & 2004 & 2005 Q_{n} (m^{3} / s) & 430 & 450 & 320 & 310 & 290 R a n k i n g & 1 & 2 & 3 & 4 & 5 \end{matrix}$
$∴ P = \frac{m}{n + 1} = \frac{3}{5 + 1}$
=0.5 or 50%

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