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Let's analyze the problem step by step:

### Given Information:
- Sample size (\(n\)) = 5
- Mean (\(\bar{x}\)) = 76632 lb/pulg
- Standard deviation (\(s\)) = 152.82 lb/pulg
- Minimum value (\(x_{\text{min}}\)) = 76537 lb/pulg
- Maximum value (\(x_{\text{max}}\)) = 76901 lb/pulg
- Median (\(\tilde{x}\)) = 76567 lb/pulg

### Problem Breakdown:
a. **Determine the values of the two sample observations.**

   Since the sample size \(n = 5\), we have five observations. The exact values are not provided, but we can infer some information from the statistics given.

b. **Calculate \(Q1\) and \(Q2\).**

   \(Q1\) is the first quartile (25th percentile), and \(Q2\) is the second quartile (50th percentile), which is also the median. The median is already given as 76567.

c. **Calculate the IQR.**

   IQR (Interquartile Range) is calculated as:
   \[
   \text{IQR} = Q3 - Q1
   \]
   where \(Q3\) is the third quartile (75th percentile).

d. **Calculate the range.**

   The range is calculated as:
   \[
   \text{Range} = x_{\text{max}} - x_{\text{min}}
   \]
   where \(x_{\text{max}} = 76901\) and \(x_{\text{min}} = 76537\).

e. **Calculate the coefficient of variation.**

   The coefficient of variation (CV) is calculated as:
   \[
   \text{CV} = \frac{s}{\bar{x}} \times 100
   \]
   where \(s\) is the standard deviation and \(\bar{x}\) is the mean.

Let's proceed with the calculations:

### Step 1: Calculate the Range
\[
\text{Range} = 76901 - 76537 = 364 \text{ lb/pulg}
\]

### Step 2: Calculate the Coefficient of Variation
\[
\text{CV} = \frac{152.82}{76632} \times 100 \approx 0.1994\%
\]

### Step 3: Determine Q1 and Q3 (Assumption)
To find \(Q1\) and \(Q3\), we typically need the actual data points or a method to estimate these values from the summary statistics provided. With a sample of 5 data points, estimating quartiles directly from the given information without raw data involves assumptions. 

For this example, assume:
- The data might be evenly distributed. This implies \(Q1\) is somewhere between the minimum value and the median, and \(Q3\) is between the median and the maximum value.

### Step 4: Calculate the IQR
Given assumptions, we can approximate \(Q1\) and \(Q3\) using the median:
- If the data is symmetrically distributed:
  \[
  Q1 \approx \frac{76537 + 76567}{2} = 76552 \text{ lb/pulg}
  \]
  \[
  Q3 \approx \frac{76567 + 76901}{2} = 76734 \text{ lb/pulg}
  \]
  Then:
  \[
  \text{IQR} = 76734 - 76552 = 182 \text{ lb/pulg}
  \]

### Summary of Results:
- **Range:** 364 lb/pulg
- **CV:** 0.1994%
- **IQR:** 182 lb/pulg
- **\(Q1\):** 76552 lb/pulg (approximation)
- **\(Q3\):** 76734 lb/pulg (approximation)

Would you like more details on these steps, or do you have any questions?

Here are some related questions you might find interesting:
1. How does the assumption of symmetry in data distribution affect the calculation of quartiles?
2. What are some methods for estimating quartiles if the actual data points are unavailable?
3. How would you interpret a low coefficient of variation in a practical scenario?
4. Why is the range considered a less robust measure of dispersion compared to the IQR?
5. How can the IQR be used to detect outliers in a dataset?

**Tip:** Understanding the relationship between different measures of dispersion (like range, IQR, and CV) can provide deeper insights into the variability and consistency of data.

Learn how to analyze statistical data using measures such as range, coefficient of variation (CV), and interquartile range (IQR). This problem involves calculating these measures from given sample statistics: mean, standard deviation, minimum, maximum, and median. Suitable for advanced high school and undergraduate students.

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