We have the following annual rainfall amounts from historic data:YearNo.12345678910RainfallIntensity,in.43.3053.0263.5245.9348.2650.5149.5743.9346.7759.12YearNo.11121314151617181920RainfallIntensity,in.54.4947.3840.7845.0550.3754.9151.2839.9153.2967.59YearNo.212223242526272829Plot the following annual rainfall amounts on probability paper as follows:a) Normal paperb) Lognormal paperc) Weibull paperd) Which distribution is best, based on the probability paper plots?RainfallIntensity,in.58.7142.9655.7741.3158.8348.2144.6767.7243.11

Let us consider the rainfall intensity amounts as X. Consider, the mean and standard deviation of X, μ=43.3+53.02+...+44.67+67.72+43.1129 =50.7 σ=∑X-X¯2n-1 =43.3-50.6992+53.02-50.6992+....+43.11-50.699229-1 =7.71 a) Normal plot: PDF of normal distribution: fx=1σ2πe-12x-μσ2 By substituting the values of μ and σ, the pdf becomes, fx=17.712πe-12x-50.77.712 =0.0517e-12x-50.77.712 X f(x) 43.3 f43.3=0.0517e-12x-50.77.712 =0.03266 53.02 0.04947 63.52 0.01299 45.93 0.04274 48.26 0.04923 50.51 0.05174 49.57 0.05121 43.93 0.0352 46.77 0.04545 59.12 0.02851 54.49 0.04587 47.38 0.04718 40.78 0.02262 45.05 0.03957 50.37 0.05171 54.91 0.04459 51.28 0.05161 39.91 0.01944 53.29 0.04892 67.59 0.0047 58.71 0.03017 42.96 0.03127 55.77 0.0417 41.31 0.02465 58.83 0.02969 48.21 0.04913 44.67 0.03812 67.72 0.00453 43.11 0.03188 Plot: Excel Procedure: Enter x and f(x) in Excel sheet Select the columns of x and f(x) Go to Insert menu. Click on “Insert scatter or bubble chart” option. Select Scatter under Charts. Under chart elements, check Trend line as Polynomial. Select 'Display -squared value on chart'. Excel Output: From the plot, the percentage of variation explained by the normal model is 0.9091=90.91%. b) Log-normal distribution: The pdf of X is, fx=1xσ2πe-12σ2lnx-μ2 Here, EX=eμ+σ2250.7=eμ+σ223.9259=μ+σ227.8518=2μ+σ2...........(1) VX=eσ2-1e2μ+σ27.712=eσ2-1*e7.85180.0231=eσ2-11.0231=eσ2σ2=0.0228 From the equation 1, 7.8518=2μ+0.02283.9145=μ By substituting the values of μ and σ, the pdf becomes, fx=1x*0.02282πe-12*0.0228lnx-3.91452 =0.0456xe-21.93lnx-3.91452 X f(x) 43.3 0.00066 53.02 0.0008 63.52 0.00021 45.93 0.00084 48.26 0.00092 50.51 0.0009 49.57 0.00092 43.93 0.00071 46.77 0.00088 59.12 0.00042 54.49 0.00072 47.38 0.0009 40.78 0.00044 45.05 0.00079 50.37 0.0009 54.91 0.00069 51.28 0.00088 39.91 0.00037 53.29 0.00079 67.59 9.5E-05 58.71 0.00045 42.96 0.00063 55.77 0.00064 41.31 0.00049 58.83 0.00044 48.21 0.00091 44.67 0.00076 67.72 9.2E-05 43.11 0.00064 Plot: Excel Procedure: Enter x and f(x) in Excel sheet Select the columns of x and f(x) Go to Insert menu. Click on “Insert scatter or bubble chart” option. Select Scatter under Charts. Under chart elements, check Trend line as Polynomial. Select 'Display -squared value on chart'. Excel Output: From the plot, the percentage of variation explained by the lognormal model is 0.838=83.8%. c) Weibull distribution: fx=kλxλk-1e-xλk EX=λΓ1+1k50.7=λΓ1+1k................(1) VX=λ2Γ1+2k-Γ1+1k27.712=λ2Γ1+2k-50.722629.934=λ2Γ1+2k..............(2) By solving the equations (1) and (2), λ=54.07k=6.741 Therefore, the pdf is, fx=kλxλk-1e-xλk =6.74154.07x54.076.741-1e-x54.076.741 =0.125x54.075.741e-x54.076.741 X f(x) 43.3 0.027921 53.02 0.046505 63.52 0.016301 45.93 0.035117 48.26 0.040893 50.51 0.044946 49.57 0.0435 43.93 0.029648 46.77 0.037318 59.12 0.033627 54.49 0.045567 47.38 0.038843 40.78 0.021317 45.05 0.032731 50.37 0.044757 54.91 0.045029 51.28 0.045809 39.91 0.019219 53.29 0.046442 67.59 0.004992 58.71 0.03513 42.96 0.026996 55.77 0.043553 41.31 0.022648 58.83 0.034696 48.21 0.040782 44.67 0.031687 67.72 0.00476 43.11 0.027403 Plot: Excel Procedure: Enter x and f(x) in Excel sheet Select the columns of x and f(x) Go to Insert menu. Click on “Insert scatter or bubble chart” option. Select Scatter under Charts. Under chart elements, check Trend line as Polynomial. Select 'Display -squared value on chart'. Excel Output: From the plot, the percentage of variation explained by the Weibull model is 0.9699=96.99%. d) From the above three plots, the percentage of variation explained by the Weibull distribution is more compared to other two models. Hence, the best model is Weibull model. Therefore, the Weibull distribution is the best distribution for fitting the annual rainfall amounts.