Step-by-step Answer:
According to the WK encyclopedia, there are three ways to calculate the quartiles of an ODD number of data points.
For EVEN number of data points (as in your other example), all three methods give the same answers.
Consider the given set of data points, {20,40,42,72,82), the three methods give the following results:
1. TI-83 method:
Remove the middle point, and take the median of each half as the respective quartiles. For the given data set, we remove the data point 42, giving
Q1 = (20+40)/2 = 30
Q3 = (72+82)/2 = 77
2. Tukey method
Include the middle point in each half of the data and calculate the median of each half.
For the given data set, we have
Q1 = median of {20,40,42} = 40
Q3 = median of (42,72,82} = 72
This corresponds to your attempt.
3. This method is dependent on the number of data points, 4n+1 or 4n+3
It is a little more complex, but logical.
There are two cases (both ODD number of data points).
Case 1: number of data points = 4n+3 :
where n is an integer, for example, 7, 11,15,19,... n=1 for 7 data points, 4*(1)+3 = 7, and so on.
Say we take a data set {12,20,40,42,72,82,86} with 7 data points, then n=1.
The lower quartile is given by 75% of point n+1, and 25% of point n+2, or
Q1 = 75% * 20 + 25% * 40 = 15 + 10 = 25
The third quartile is given by 25% of point 3n+2 (point 5 = 72, and 75% of point 3n+3 (point 6 = 82)
Q3 = 25% * 72 + 75% * 82 = 18 + 61.5 = 79.5
Case 2: number of data points = 4n+1
Say we have a data set {20,40,42,72,82},
number of data points = 5 = 4n+1, where n=1
The lower quartile is given by 75% of data point n (Point 1 = 20) and 25% of point n+1 (point 2 = 40)
Q1 = 25%*20 + 75%*40 = 5+30 = 35
The upper quartile is given by 75% of data point 3n+1 (Point 4 = 72) and 25% of point 3n+2 (point 5 = 82)
Q3 = 75%*72 + 75% * 82 = 54 + 20.5 = 74.5
For even number of data points, all three methods give the same results for quartiles.
We see that method 3 has results lying between methods 1 and 2, both of which are over-simplifications. However, with large number of data points, the difference is minimum.
Finally, with three sets of answers, which one should we choose?
We know that you have attempted the results using method 2, and were told that they were incorrect. That leaves us with the first and the third methods.
Most schools do not get into details with method 3 (unless the two other methods have been previously introduced. This means I would bet on method 1 without certainty.
