Red · Yamhill-Carlton District · United States
Ken Wright Cellars Sedimentary Pinot Noir
Scored from 160 calibrated Vivino reviewers, each re-centered on their own rating baseline before the bottle is ranked. Cohort: Red · United States (1,156 wines).
Find this wine: view on Vivino · compare prices on Wine-Searcher
What reviewers say
“A round and robust Pinot, nose chalk-full of caramelized red fruit and fresh rose petal aromas.”
From a top-rated Vivino review for this bottle.
Summary
Red plum, fresh fig and marionberry fruits with hints of magnolia blossom and graham cracker. Balanced but vibrant acidity.
Ken Wright Cellars Sedimentary Pinot Noir is an American red from Yamhill-Carlton District. At $45.83 for the lowest 750 ml listing, it belongs to the $30–60 band.
The calibrated figure is built from 160 reviewers whose rating history varies enough to place their own average. A broader pool of 160 reviewers with two or more ratings feeds the raw-average track instead. Its in-cohort standing is measured against 1,155 other reds from the United States, not against the corpus as a whole.
How the score was built
We compare the reviewers of this bottle against every other wine those same people reviewed. Each review is scored relative to the reviewer's own baseline — four stars from someone who hands out fives means something very different from four stars from someone who rarely goes above three. Every reviewer then counts the same, so the result is a calibrated consensus rather than a star average.
- Global percentile
- Where Ken Wright Cellars Sedimentary Pinot Noir lands against every wine we rank.
- In-cohort percentile
- The same comparison narrowed to its cohort — wines of the same type from the same country. A Chilean red is judged against Chilean reds, not against Champagne. Ranked against Red · United States (1,156 wines).
- AI-adjusted percentile
- The same reviewer-normalized score, corrected for how much confidence the sample supports: thin samples are pulled toward the corpus average. Each wine keeps the share n / (n + 53) of its measured distance from the average, where n is its calibrated-reviewer count, and the result is ranked on the same scale as the global percentile.53 is the median sample size across everything we rank. Here n is 160.
