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
database name:db_id mental_health_survey
Question table Question Cols: question text dtype text, question id dtype integer, 
Survey table Survey Cols: Survey ID dtype integer, Description dtype text, 
Answer table Answer Cols: Answer Text dtype text, Survey ID dtype integer, User ID dtype integer, Question ID dtype integer, 
