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I am trying out a formula I saw in google but its not working in my excel and producing an error. Can you please help me? =(NETWORKDAYS(StartDT,EndDT)-1)*(StopTime-StartTime)+IF(NETWORKDAYS(EndDT,EndDT),MEDIAN(MOD(EndDT,1),StopTime,StartTime),StopTime)-MEDIAN(NETWORKDAYS(StartDT,StartDT)*MOD(StartDT,1),StopTime,StartTime)
Solved by T. U. in 29 mins
Hi, my name is Henry. I am working with the excel to analyse my data for psychological experiment. I have to code 4 condition. 0=both masculinity score and femininity score lower than median 1=masculinity score over median but feminity score lower than median, and it is male 2= masculinity score lower than median but femininity score over median, and it is female 3=both score over median I tried to make the "nested IF function" but it said error after I made this formula: =IF((AX>=2.727272,AY>=3.117646),"3", IF((AU=1,AX>=2.727272,AY<3.117647),"1", IF((AU=2,AX<2.727272,AY>=3.117647),"2","0"))) I dont get why it is wrong, thanks!
Solved by C. S. in 25 mins
I have a pivot table and would like to incorporate a median calculation along with the average calc. Avg is easy as it's one of the calculations available in the list but not median.
Solved by S. D. in 21 mins
I'm looking for a formula that will help me find anomalies in my Median column if the calculated value is 35% above the median then consider it as an anomaly
Solved by M. Q. in 19 mins
I need to find the median off a list of numbers that match a criteria of the row next to it that are percents. the median needs to be based of the numbers that fall in between 2 different percents
Solved by K. D. in 19 mins