Calculate 90th continuous percentile of each region | SQL | SQL Interview question

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[@srivalli5269]

Well explained without predefined functions

[@sahitikarnatakapu2829]

Great sir.. You are intelligent.

[@vijaygupta7059]

Thanks for new function, I have learn new function today

[@datasculptor2895]

Glad to hear that

[@Factstelugu5]

sql intemediate advanced topics videos/resources cheyyandi bro

[@datasculptor2895]

Sure.. definitely. Please subscribe to my channel.

[@sravankumar1767]

SELECT Region, PERCENTILE_CONT(0.9) WITHIN GROUP (ORDER BY Amount) OVER (PARTITION BY Region) AS percentile_90 FROM table_name;

[@datasculptor2895]

Yes. Same is mentioned in the video.

[@sravankumar1767]

SELECT Region, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY Amount) OVER (PARTITION BY Region) AS median_amount FROM Sales;

[@datasculptor2895]

Yes. Same is mentioned in the video.

[@UnrealAdi]

MySQL: WITH RankedData AS ( SELECT Region, revenue, ROW_NUMBER() OVER (PARTITION BY Region ORDER BY revenue) AS rn, COUNT(*) OVER (PARTITION BY Region) AS cnt FROM sales2 ), PercentilePositions AS ( SELECT Region, revenue, rn, cnt, 0.9 * (cnt - 1) + 1 AS percentile_pos FROM RankedData ), InterpolatedData AS ( SELECT Region, revenue AS lower_revenue, LEAD(revenue) OVER (PARTITION BY Region ORDER BY revenue) AS upper_revenue, rn, FLOOR(percentile_pos) AS lower_rn, CEIL(percentile_pos) AS upper_rn, percentile_pos FROM PercentilePositions ) SELECT *, CASE WHEN lower_rn = upper_rn THEN lower_revenue ELSE lower_revenue + (upper_revenue - lower_revenue) * (percentile_pos - lower_rn) END AS percentile_90 FROM InterpolatedData WHERE rn = lower_rn -- OR rn = upper_rn ORDER BY Region;

[@datasculptor2895]

Is this giving the correct result?