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 2 added 38 characters in body edited Dec 15 '16 at 6:42 Michael Chernick 35.8k88 gold badges6363 silver badges134134 bronze badges Take the exponential distribution with rate parameter a and density a exp(-ax) for 0<=x< infinity. The mode is at zero. Of course the mean and the median are greater than 0. The cdf is 1-exp(-ax). So for the median solve for exp(-ax)=0.5 for x. Then -ax=ln(0.5) or x = -ln(0.5)/a. For the mean integrate ax exp(-ax) from 0 to infinity. Take a=1 and we have a median = -ln(0.5)=ln(2) and mean = 1.   So mode < median < mean. Take the exponential distribution with rate parameter a and density a exp(-ax) for 0<=x< infinity. The mode is at zero. Of course the mean and the median are greater than 0. The cdf is 1-exp(-ax). So for the median solve for exp(-ax)=0.5 for x. Then -ax=ln(0.5) or x = -ln(0.5)/a. For the mean integrate ax exp(-ax) from 0 to infinity. Take a=1 and we have a median = -ln(0.5).   Take the exponential distribution with rate parameter a and density a exp(-ax) for 0<=x< infinity. The mode is at zero. Of course the mean and the median are greater than 0. The cdf is 1-exp(-ax). So for the median solve for exp(-ax)=0.5 for x. Then -ax=ln(0.5) or x = -ln(0.5)/a. For the mean integrate ax exp(-ax) from 0 to infinity. Take a=1 and we have a median = -ln(0.5)=ln(2) and mean = 1. So mode < median < mean. 1 answered Dec 15 '16 at 6:11 Michael Chernick 35.8k88 gold badges6363 silver badges134134 bronze badges Take the exponential distribution with rate parameter a and density a exp(-ax) for 0<=x< infinity. The mode is at zero. Of course the mean and the median are greater than 0. The cdf is 1-exp(-ax). So for the median solve for exp(-ax)=0.5 for x. Then -ax=ln(0.5) or x = -ln(0.5)/a. For the mean integrate ax exp(-ax) from 0 to infinity. Take a=1 and we have a median = -ln(0.5).