From af73a87c4c7fb3d8c4d334544a6b28e6d5856913 Mon Sep 17 00:00:00 2001
From: Joe Heffer <60133133+Joe-Heffer-Shef@users.noreply.github.com>
Date: Thu, 23 Jan 2025 09:13:50 +0000
Subject: [PATCH] Update episodes/02-sql-aggregation.md

Co-authored-by: James Foster <38274066+jd-foster@users.noreply.github.com>
---
 episodes/02-sql-aggregation.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/episodes/02-sql-aggregation.md b/episodes/02-sql-aggregation.md
index 91aaaec4..336c52f8 100644
--- a/episodes/02-sql-aggregation.md
+++ b/episodes/02-sql-aggregation.md
@@ -226,7 +226,7 @@ Write a query that returns, from the `species` table, the number of
 
 ## Solution
 
-This query counts the number of species records that contain each value of the `taxa` field and names that result `species_count`.
+This query counts the number of times each value (Bird, Rabbit, Reptile or Rodent) in the `taxa` field occurs, defining a new field named `taxa_count` to hold the result.
 The `GROUP BY` clause means the query will create an aggregated table with one row for each taxa.
 Only those `taxa` values that have more than ten records will be included because of the `HAVING` clause.
 This filtering is applied _after_ grouping has been done.