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Yearly peak flows Freqanaly River measured at a gaging site over a period of 35 years (ranked in the descending order) are gi
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Answer #1

It is assumed that data is following a normal distribution and the formulas of normal distribution is applied.

Whenever data follows log normal distribution we need to make the data into a normal data by applying natural logarithm to them.

Now, we can apply the normal dirstribution formulas and the obtained result is converted back into a result of lognormal by applying exponential to it.

G43 : =G41/37 F J K L M N P R A Rank 1 1 LV487777 2 3 4 2 3 4 5 6 7 8 9 5 6 7 8 10 11 12 13 14 15 16 17 9 10 11 12 13 Mean St

2 Q=1970-3 m/s = 33.25135 msec (Mean discharge). 1970.3 37 n 2(Q - Qrean) ² standard deviation n-1 16315-51243 37 -1 standard

EY= 143.0845 2 143.0845 37 = 3.867147 sy = (y-7) n- to allo8543 Sy = 0.527 7336742 T=100 = ý tkp sy = 3.867147+ (2.325) 6 52

log-Pearson Type II distribution CSE 0 (x-2)3 (n-1) (A-2) s} = 37 3G:-) :-54465 02084 5=2128869838 (5 = 37 (-54765.02084 (36)

«т: 2+ (-)} + 4 (2247) - (-) + 2 + 4 5 а ) + (5-ла1)(на) - (-а). to us) ояту - (2 раза) (- 9-ого4 (an (2010), (- гнаў , 2-126

Chipboard Font Alignment Number Styles C1 (x-xmean) F J K L M 1 vunna/ Mean Standard Error Median Mode Standard Deviation Sam

Calculations are carried out in Ms excel for accuracy and for easy computating facility.

All the formulas are given in the spread sheet image.Kindly use them.

I hope you have understood.

All the Best.!

Please do mention your questions in the comments seciton.

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