Quartiles

a) We can find the position of the lower quartile in the data set with the formula

Q_1 Position = \dfrac{n+1}{4}

There are 19 entries in the data set and so n = 19

Q_1 Position = \dfrac{19+1}{4}=\dfrac{20}{4}=5,

The 5^{th} value in the data set is the lower quartile, which is 43.

b) We can find the position of the upper quartile in the data set with the formula

Q_3 Position = \dfrac{3(n+1)}{4}

There are 19 entries in the data set and so n = 19

Q_3 Position = \dfrac{3(19+1)}{4}

Q_3 Position = \dfrac{3 \times 20}{4} = \dfrac{60}{4} = 15

The 15^{th} value in the data set is the upper quartile, which is 168

Firstly we must order this data. Which when done gives us

13, 27, 33, 37, 55, 57, 71, 77, 84, 94, 97

The data set has 11 elements. So we can find the position of Q_1 and Q_3. 

Q_1 Position = \dfrac{11+1}{4} = 3

The third element is 33 and so it is the lower quartile.

Q_3 Position = \dfrac{3(11+1)}{4} = 9

The ninth element is 84 and so it is the upper quartile.

Then the interquartile range

IQR = Q_3 -Q_1
IQR = 84 - 33 = 51

Both classes 1 and 2 have 11 students. So for both, the lower quartile is in the 3^{\text{rd}} position and the upper quartile is in the 9^{\text{th}} position.

Class 1:
Q_3 =99
Q_1 = 72
IQR = 99 - 72 = 27

Class 2:
Q_3 =79
Q_1 = 71
IQR = 79 - 71 = 8

While class 2 had the lower median grade, it did have the more consistent grades, despite outliers.