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How have American baby name tastes changed since 1920? Which names have remained popular for over 100 years, and how do those names compare to more recent top baby names? These are considerations for many new parents, but the skills you'll practice while answering these queries are broadly applicable. After all, understanding trends and popularity is important for many businesses, too!

You'll be working with data provided by the United States Social Security Administration, which lists first names along with the number and sex of babies they were given to in each year. For processing speed purposes, the dataset is limited to first names which were given to over 5,000 American babies in a given year. The data spans 101 years, from 1920 through 2020.

The Data

baby_names

columntypedescription
yearintyear
first_namevarcharfirst name
sexvarcharsex of babies given first_name
numintnumber of babies of sex given first_name in that year
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DataFrameas
usa_baby_names
variable
-- Run this code to view the data in baby_names
SELECT *
FROM baby_names;
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DataFrameas
name_types
variable
-- Use this table for the answer to question 1:
SELECT first_name, SUM(num) AS sum, 
	CASE WHEN first_name >= '50' THEN 'Classic'
 	ELSE 'Trendy' END AS popularity_type
FROM baby_names
GROUP BY first_name, popularity_type
ORDER BY first_name ASC
LIMIT 5;
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DataFrameas
top_20
variable
-- Use this table for the answer to question 2:
WITH ranked_names AS (
    SELECT 
        first_name,
        SUM(num) AS sum,
        RANK() OVER (ORDER BY SUM(num) DESC) AS name_rank
    FROM baby_names
    WHERE sex = 'M'
    GROUP BY first_name
)
SELECT 
    name_rank,
    first_name,
    sum
FROM ranked_names
WHERE name_rank <= 20 
   OR first_name = 'Paul'
ORDER BY name_rank;
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DataFrameas
a_names
variable
-- Use this table for the answer to question 3:
SELECT a.first_name, SUM(a.num + b.num) AS total_occurrences
FROM baby_names AS a
JOIN baby_names AS b
ON a.first_name = b.first_name
WHERE a.year = '1920' AND b.year = '2020' AND a.sex = 'F' AND b.sex = 'F'
GROUP BY a.first_name;