Among respondents who participated in the survey in 2016, what percentage had a mental health disorder in the past?
respondents and 'users' are synonyms; percentage = divide(count(SurveyID = 2016& QuestionID = 32 & AnswerText = 'Yes'), count(SurveyID = 2016& QuestionID = 32))*100%
SELECT CAST(SUM(CASE  WHEN T1.AnswerText LIKE 'Yes' THEN 1 ELSE 0 END) AS REAL) * 100 / COUNT(T1.SurveyID) FROM Answer AS T1 INNER JOIN Question AS T2 ON T1.QuestionID = T2.questionid WHERE T1.QuestionID = 32 AND T1.SurveyID = 2016