The IQR (interquartile range) is the difference between Q3 and Q1.

For the data below (same data as in previous question) find the IQR:

IQR=

A value is considered unusually small if it is 1.5*IQR smaller then the median and is considered unusually large if it is 1.5*IQR larger then the median.

Based on this, what is the smallest usual value for the data set? (do not round, type as a decimal)

What is the largest usual value for the data set? (do not round, type as a decimal)

An outlier would be a data value which is smaller then the smallest usual data value or larger then the largest data value. If there is an outlier, type the value of the outlier below. Otherwise type N.

Data set:

5
5
5
5
5
6
7
7
8
6
7
5
3
5
8
7
9
1
2
3
4
5
3
5
8
7
9
6
3
5
2
4
5

Let's go step by step to solve this problem.

### 1. **Arrange the Data Set in Ascending Order:**
First, we need to arrange the data set in ascending order.

\[
\text{Data Set: } 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 9, 9
\]

### 2. **Find Q1, Q3, and IQR:**
- **Q1 (First Quartile):** The median of the first half of the data.
- **Q3 (Third Quartile):** The median of the second half of the data.
- **IQR:** The difference between Q3 and Q1.

#### Finding Q1:
The data set has 31 values. The first half contains 15 values. Q1 is the median of the first 15 values:

\[
\text{First half: } 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5
\]
The median (8th value): Q1 = 4

#### Finding Q3:
The second half contains 15 values. Q3 is the median of the last 15 values:

\[
\text{Second half: } 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 9
\]
The median (8th value): Q3 = 7

#### Calculating IQR:
\[
\text{IQR} = Q3 - Q1 = 7 - 4 = 3
\]

### 3. **Determine the Smallest and Largest Usual Values:**
- **Smallest Usual Value:** \( \text{Median} - 1.5 \times \text{IQR} \)
- **Largest Usual Value:** \( \text{Median} + 1.5 \times \text{IQR} \)

#### Median of the Data Set:
The median is the middle value of the entire data set, which is the 16th value:
\[
\text{Median} = 5
\]

#### Smallest Usual Value:
\[
\text{Smallest Usual Value} = 5 - 1.5 \times 3 = 5 - 4.5 = 0.5
\]

#### Largest Usual Value:
\[
\text{Largest Usual Value} = 5 + 1.5 \times 3 = 5 + 4.5 = 9.5
\]

### 4. **Identify Outliers:**
- Any value smaller than 0.5 or larger than 9.5 would be considered an outlier.

#### Checking the Data Set:
- The smallest value in the data set is 1, which is greater than 0.5.
- The largest value in the data set is 9, which is less than 9.5.

So, there are **no outliers**.

### Final Answers:
- **IQR** = 3
- **Smallest Usual Value** = 0.5
- **Largest Usual Value** = 9.5
- **Outlier** = N (no outliers)

Would you like further details or have any questions?

Here are 5 additional questions to expand on this topic:
1. How would the IQR change if the largest value in the data set were 12 instead of 9?
2. What is the significance of using the IQR in determining outliers?
3. How would the result change if you used the mean and standard deviation to identify outliers?
4. Can you explain how skewness affects the median, quartiles, and IQR?
5. What are the limitations of using IQR for detecting outliers?

**Tip:** The IQR is a measure of statistical dispersion and is particularly useful in identifying outliers in a data set that is not normally distributed.

Learn how to calculate the interquartile range (IQR) for a given data set and determine usual values based on the median and IQR. This problem explores descriptive statistics concepts, including quartiles and outlier detection, suitable for Grades 7-9.

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