Math  /  Data & Statistics

QuestionConsider the below data for ALL PARTS of this question: 68322859454752554750\begin{array}{llllllll}68 & 32 & 28 & 59 & 45 & 47 & 52 & 55 \\ 47 & 50\end{array} Enter an OUTLIER value if any or NONE. \qquad
Note: Enter X.X AT LEAST ONE DIGIT BEFORE THE DECIMAL, ONE AFTER and round up AFTER all calculations. Thus, 7 is entered as 7.0;3.5627.0 ;-3.562 is entered as 0.3750 is entered as 0.4;17.3510.4 ; 17.351 is entered as 17.4 \square A Answer 5

Studdy Solution

STEP 1

What is this asking? We need to find any outlier in a given set of numbers. Watch out! Remember, an outlier is a number that's way different from all the others.
Don't just pick the biggest or smallest number – it has to be *significantly* bigger or smaller.

STEP 2

1. Sort the data
2. Calculate Quartiles
3. Calculate IQR
4. Calculate Bounds
5. Find Outliers

STEP 3

Let's **sort** our data from smallest to largest.
This helps us see the spread and find those quartiles easily!
Our sorted data is: 28,32,45,47,47,50,52,55,59,6828, 32, 45, 47, 47, 50, 52, 55, 59, 68.

STEP 4

Now, let's find the **quartiles**!
The **second quartile (Q2)** is the median of the whole dataset.
Since we have 10\bf{10} numbers (an even amount), the median is the average of the 5th\bf{5^{th}} and 6th\bf{6^{th}} numbers.
So, Q2 = 47+502=972=48.5\frac{47 + 50}{2} = \frac{97}{2} = 48.5.

STEP 5

The **first quartile (Q1)** is the median of the lower half of the data.
The lower half is 28,32,45,47,4728, 32, 45, 47, 47.
The middle number is 45\bf{45}, so Q1 = 4545.

STEP 6

The **third quartile (Q3)** is the median of the upper half of the data.
The upper half is 50,52,55,59,6850, 52, 55, 59, 68.
The middle number is 55\bf{55}, so Q3 = 5555.

STEP 7

The **Interquartile Range (IQR)** is the difference between Q3 and Q1.
It tells us how spread out the middle 50% of our data is.
So, IQR = Q3 - Q1 = 5545=1055 - 45 = \bf{10}.

STEP 8

To find potential outliers, we calculate the **lower and upper bounds**.
The **lower bound** is Q1 - 1.5 * IQR = 451.510=4515=3045 - 1.5 \cdot 10 = 45 - 15 = \bf{30}.
The **upper bound** is Q3 + 1.5 * IQR = 55+1.510=55+15=7055 + 1.5 \cdot 10 = 55 + 15 = \bf{70}.

STEP 9

Any number outside these bounds is a potential outlier.
Looking back at our sorted data (28,32,45,47,47,50,52,55,59,6828, 32, 45, 47, 47, 50, 52, 55, 59, 68), we see that 28\bf{28} is less than our lower bound of 30\bf{30}.
All other numbers are within our calculated bounds.

STEP 10

The outlier is 28.028.0.

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