Math  /  Data & Statistics

QuestionA set of data comprises of five numbers x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5} which have been placed in ascending order. a. Recalling definitions, such as the Lower Quartile is the n+14th\frac{n+1}{4} t h piece of data with the data placed in order, find an expression for the Interquartile [2]
Range. b. Hence, show that a data set with only 5 numbers in it cannot have any outliers. [5] c. Give an example of a set of data with 7 numbers in it that does have an outlier, justify this fact by stating the Interquartile Range. [2]

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the interquartile range of five ordered numbers, prove that this set can't have outliers, and then give an example of a seven-number set *with* an outlier. Watch out! Remember outliers are values significantly above or below the "normal" range, specifically outside 1.5 times the interquartile range from the quartiles.
Don't mix up the upper and lower quartiles!

STEP 2

1. Find the Interquartile Range Formula
2. Prove No Outliers with Five Numbers
3. Example of a Seven-Number Set with an Outlier

STEP 3

Alright, let's **define** our five ascending numbers as x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5.
Since they're already in order, finding the quartiles is a piece of cake!

STEP 4

The **lower quartile** (Q1Q_1) is the n+14\frac{n+1}{4}th value.
With n=5n = 5, that's 5+14=64=32=1.5\frac{5+1}{4} = \frac{6}{4} = \frac{3}{2} = 1.5.
So, Q1Q_1 is halfway between the **first** and **second** numbers, x1x_1 and x2x_2.
We can write this as Q1=x1+x22Q_1 = \frac{x_1 + x_2}{2}.

STEP 5

Similarly, the **upper quartile** (Q3Q_3) is the 3(n+1)4\frac{3(n+1)}{4}th value.
Plugging in n=5n = 5, we get 3(5+1)4=184=92=4.5\frac{3(5+1)}{4} = \frac{18}{4} = \frac{9}{2} = 4.5.
This means Q3Q_3 is halfway between the **fourth** and **fifth** numbers, x4x_4 and x5x_5.
We write this as Q3=x4+x52Q_3 = \frac{x_4 + x_5}{2}.

STEP 6

The **interquartile range (IQR)** is just the difference between the upper and lower quartiles: IQR=Q3Q1=x4+x52x1+x22=x4+x5x1x22IQR = Q_3 - Q_1 = \frac{x_4 + x_5}{2} - \frac{x_1 + x_2}{2} = \frac{x_4 + x_5 - x_1 - x_2}{2}

STEP 7

Let's think about what an outlier *is*.
A value is an outlier if it's more than 1.5IQR1.5 \cdot IQR away from the nearest quartile.

STEP 8

For the **smallest value**, x1x_1, to be an outlier, it would need to be less than Q11.5IQRQ_1 - 1.5 \cdot IQR.
Let's see if that's possible. Q11.5IQR=x1+x221.5x4+x5x1x22Q_1 - 1.5 \cdot IQR = \frac{x_1 + x_2}{2} - 1.5 \cdot \frac{x_4 + x_5 - x_1 - x_2}{2} =2x1+2x21.5x41.5x5+1.5x1+1.5x22=3.5x1+3.5x21.5x41.5x52= \frac{2x_1 + 2x_2 - 1.5x_4 - 1.5x_5 + 1.5x_1 + 1.5x_2}{2} = \frac{3.5x_1 + 3.5x_2 - 1.5x_4 - 1.5x_5}{2}Since x1x2x_1 \le x_2 and x4x5x_4 \le x_5, this value is always greater than or equal to x1x_1.
Thus, x1x_1 cannot be an outlier.

STEP 9

Similarly, for the **largest value**, x5x_5, to be an outlier, it would need to be greater than Q3+1.5IQRQ_3 + 1.5 \cdot IQR. Q3+1.5IQR=x4+x52+1.5x4+x5x1x22Q_3 + 1.5 \cdot IQR = \frac{x_4 + x_5}{2} + 1.5 \cdot \frac{x_4 + x_5 - x_1 - x_2}{2} =2x4+2x5+1.5x4+1.5x51.5x11.5x22=3.5x4+3.5x51.5x11.5x22= \frac{2x_4 + 2x_5 + 1.5x_4 + 1.5x_5 - 1.5x_1 - 1.5x_2}{2} = \frac{3.5x_4 + 3.5x_5 - 1.5x_1 - 1.5x_2}{2}Since x1x2x_1 \le x_2 and x4x5x_4 \le x_5, this value is always less than or equal to x5x_5.
Thus, x5x_5 cannot be an outlier.

STEP 10

Consider the set {1, 2, 3, 4, 5, 6, 15}. Q1Q_1 is the second value (2), and Q3Q_3 is the sixth value (6).
The IQR is 62=46 - 2 = 4.

STEP 11

An outlier would be any value greater than 6+1.54=126 + 1.5 \cdot 4 = 12 or less than 21.54=42 - 1.5 \cdot 4 = -4.
Since 15>1215 > 12, 15\mathbf{15} is an **outlier**!

STEP 12

a. The interquartile range for five ordered numbers is x4+x5x1x22\frac{x_4 + x_5 - x_1 - x_2}{2}. b. We proved that neither the smallest nor largest value can be outliers in a five-number set. c. The set {1, 2, 3, 4, 5, 6, 15} has an outlier at 15, since the IQR is 4 and 15 is more than 1.5 * 4 away from the upper quartile.

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