GROUP BY CASE 
         WHEN VALUE >= 10 AND VALUE <= 20 THEN '20%'
         WHEN VALUE > 20 AND VALUE <= 32.5 THEN '40%' ELSE '0' END

You need to cover all possible values, hence the ELSE 0. You'll probably want to do something a little different there, but it should give you a start.

Based on Joel Gauvreau's comment:

SUM(CASE WHEN VALUE >=10 AND VALUE <= 20 THEN 1.0 ELSE 0.0 END) / COUNT(*),
SUM(CASE WHEN VALUE > 20 AND VALUE <= 32.5 THEN 1.0 ELSE 0.0 END) / COUNT(*)

Or at the end of the query use the COMPUTE statement.

SELECT B.Description, Total = COUNT(*) / CONVERT(money, (SELECT COUNT(*) FROM Target T2))
FROM Target T
JOIN (
    SELECT  Description = '0 to 10', LBound = 0, UBound = 10 
    UNION ALL 
    SELECT Description = '10 to 20', LBound = 10, UBound = 20
) B ON T.Value >= LBound AND T.Value < B.UBound
GROUP BY B.Description

This will get you the count per range, you can easily determine the percentage from there:

declare @ranges table (beginInclusive float, endExclusive float)
insert @ranges (beginInclusive, endExclusive)
    select 10, 20
    union all select 20, 32.5
select
    r.beginInclusive,
    r.endExclusive,
    count(*)
from t join @ranges on t.RangedValue >= r.beginInclusive and t.RangedValue < r.endExclusive
group by 
    r.beginInclusive,
    r.endExclusive

Joel's answer seems the best way to me. Posting to explain the query, and because the answer has an integer division sum/count which will return 1 or 0 instead of a percentage.

For the 20 -> 32.5 range:

select CAST(SUM(
    CASE WHEN 20 < field AND field <= 32.5 THEN 1 ELSE 0 END
    ) as float) / COUNT(*) * 100.0
from table

The case returns 1 when the value is in range. Because there's no group by clause, the sum adds the result of the case for every row in the table. Convert to float, divide by the number of rows =count(*), and you get the percentage.

You can also write it like:

select SUM(
    CASE WHEN 20 < field AND field <= 32.5 THEN 1.0 ELSE 0.0 END
    ) / COUNT(*) * 100
from table

Here the CASE will result in a float 1.0 instead of the integer 1.

I would usually use a subquery and get rangecounts and join in the total to get the percentage. Something like:

SELECT 
  RangeCount/CNT as Percentage,
  Range
FROM 
(
SELECT
  Count(*) AS CNT
FROM
  SomeTable
) AS Total
LEFT JOIN 
(
SELECT
  CASE Val <= 10 then
       '0 up to 10'
  ELSE 
       CASE when Val <= 20
         '11 to 20'
       ELSE 
        '> 20'
       END
    END
  END AS Range,
  COUNT(*) AS RangeCount
FROM 
   SomeTable
GROUP BY
   Range
) AS RangeTotals

Declare @1 as int
Declare @2 as int
Declare @TotalRows as int
set @1 = (Select COUNT(id) FROM dbo.Table_1 WHERE id >= 10 and id <= 20)
set @2 = (Select COUNT(id) FROM dbo.Table_1 WHERE id > 20 AND id <= 32.5);
set @TotalRows = (Select Count(id) from dbo.Table_1);
SELECT CAST(((@1 * 100)/@TotalRows) as nvarchar(32)) + '%', CAST(((@2 * 100)/@TotalRows) as nvarchar(32)) + '%';

Little complicated, but that does work... i suppose...

dbo.Table_1 only has 1 column, 'id' and it is of type int.

If it's something that you will be doing regularly, then you can create a table with the ranges and what constitutes those ranges. If not, you can set them up in a table variable or temporary table and join to that. It's basically JohnOpincar's solution, but with a table instead of a subquery.

Also, in your example you list "10 to 20" and "20 to 32.5". Where is a row counted if it's exactly 20? You should probably make sure that your requirements are clear on that point before you deliver the final solution.

Readers who found this question after a Google search are most likely, like me, wanting to create a continuous range.

Using the FLOOR() method, you can mathematically create this range with minimal code:

SELECT 
    COUNT(*) AS patients_in_group, 
    -- Creates groups in ranges of 10 (60, 70, 80, etc.)
    FLOOR(weight/10) * 10 AS weight_group
FROM patients
GROUP BY weight_group

For the specific case of this question, where ranges are not continuous, the previous answer by Joel Coehoorn, using CASES, is the most adequate. It is easy to read and understand.