Are you kidding? Ugly is my middle name... 😛

I don't think the value in bold is correct. If I understand your computation correctly you're trying to find the median of the lower half of the grades for 2011, which would be in this set from your input stream:

114 69.0 2011-03-02

115 70.0 2011-03-02

116 75.0 2011-03-02

118 76.0 2011-03-02

113 76.0 2011-03-02

And that value is 116's score of 75.

So here is my solution, my ugly baby. Gotta love it cause it's my baby!

;WITH [Stats] AS (

SELECT PersonID, Grade, CourseDate, [Year]

,Median=CASE WHEN Count1%2 = 0 AND rn1 IN (Count1/2, (Count1/2)+1) THEN Grade

WHEN Count1%2 = 1 AND rn1 = (Count1/2)+1 THEN Grade END

-- Used to establish median over quartiles

,rn2=ROW_NUMBER() OVER (PARTITION BY [Year], Quartile ORDER BY Grade)

,Count2=COUNT(*) OVER (PARTITION BY [Year], Quartile)

,Quartile

FROM (

SELECT PersonID, Grade=CAST(Grade AS DECIMAL(10,1)), CourseDate, [Year]

-- Used to establish median over years

,rn1=ROW_NUMBER() OVER (PARTITION BY [Year] ORDER BY Grade)

,Count1=COUNT(*) OVER (PARTITION BY [Year])

-- Break the grades into two groups to establish "median" in a quartile

,Quartile=NTILE(2) OVER (PARTITION BY [Year] ORDER BY Grade)

FROM EStats

CROSS APPLY (SELECT [Year]=LEFT(CourseDate, 4)) a) a

)

SELECT [Year]

,Median=AVG(Median)

,[Max]=MAX(Grade)

,[Min]=MIN(Grade)

,LowerQ=AVG(CASE

WHEN Quartile = 1 AND Count2%2 = 0 AND rn2 IN (Count2/2, (Count2/2)+1) THEN Grade

WHEN Quartile = 1 AND Count2%2 = 1 AND rn2 = (Count2/2)+1 THEN Grade END)

,UpperQ=AVG(CASE

WHEN Quartile = 2 AND Count2%2 = 0 AND rn2 IN (Count2/2, (Count2/2)+1) THEN Grade

WHEN Quartile = 2 AND Count2%2 = 1 AND rn2 = (Count2/2)+1 THEN Grade END)

FROM [Stats]

GROUP BY [Year]

Let me know if this helps.

Edit: Tidied up my solution a bit.