Q1 = 35.75
Q2 = 40
Q3= 45.5
IQR = 9.75
Lower Outlier =15
Upper Outlier=55
1) Let's calculate the quartiles, by using a formula for that and considering that the Distributions is:
2) But we need to orderly write this distribution, so:
15 29 29 35 35 36 36 37 38 40 40 42 44 45 45 47 51 52 52 55
The first Quartile is given by the formula, below where n is the number of observations in this case, since we have a decimal let's find the average between the 5th and the 6th number:
[tex]\begin{gathered} Q_1=\frac{1}{4}(n+1)^{th} \\ Q_1=\frac{1}{4}(20+1)^{th} \\ Q_1=\frac{1}{4}(21)^{th} \\ Q_1=5.25 \\ Q_{,1}=\frac{35+36}{2}=35.75 \end{gathered}[/tex]
Then The upper Quartile:
[tex]\begin{gathered} Q_3=\frac{3}{4}(n+1)^{th} \\ Q_3=\frac{3}{4}(21)^{th} \\ Q_3=\text{ 15.75 position} \\ Q_3=\frac{45+47}{2}=45.5 \end{gathered}[/tex]
3) And the Second Quartile is going to be the median
[tex]Q_2=\text{ Median =}\frac{40+40}{2}=40[/tex]
The interquartile range is going to be the difference, between the first quartile and the third one
IQR = 45.5 -35.75 =9. 75
The outliers in the distribution
15 29 29 35 35 36 36 37 38 40 40 42 44 45 45 47 51 52 52 55
They can be found by a formula:
[tex]\begin{gathered} Lower\colon Q_1-(1.5\text{ }\times IQR) \\ \text{Lower: 35.75-(1.5}\times9.75) \\ L=35.75-(14.625) \\ L=21.125\approx21 \\ \\ \text{Upper: Q}_3+(1.5\times IQR) \\ \text{Upper: }45.5+(1.5\times9.75) \\ \text{Upper: }60.125\approx60 \end{gathered}[/tex]
The lower outlier is below 21.125, and the upper one 60.125 so in our distribution, Lowe Outlier is 15, and the Upper one, is closer to 60.125 in this case, 55.
The answers are:
Q1 = 35.75
Q2 = 40
Q3= 45.5
IQR = 9.75
