2) The aggregate used in road construction begins degrading integrity of said roadway when pieces begin exceeding 23.5 millimeters in diameter. Any pieces smaller than 23.5mm in diameter are fine. The recently delivered load of aggregate being used has a 19mm mean and 5.5mm standard deviation. What percent of the load is usable if we set the max size as 23mm?
2) The aggregate used in road construction begins degrading integrity of said roadway when pieces begin exceeding 23.5 millimeters in diameter. Any pieces smaller than 23.5mm in diameter are fine. The recently delivered load of aggregate being used has a 19mm mean and 5.5mm standard deviation. What percent of the load is usable if we set the max size as 23mm?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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2. Please help me answer this including the z-table.
![### Understanding Aggregate Size in Road Construction
---
#### Problem Statement:
The aggregate used in road construction begins to degrade the integrity of the roadway when pieces exceed 23.5 millimeters in diameter. Any pieces smaller than 23.5 millimeters in diameter are considered acceptable.
The recently delivered load of aggregate has a mean size of 19 millimeters and a standard deviation of 5.5 millimeters.
> **Question:**
> What percent of the load is usable if we set the maximum allowable size as 23 millimeters?
---
To determine the usable percentage of the aggregate load, we will need to use the principles of normal distribution.
### Steps to Solve the Problem:
1. **Identify the Given Data:**
- Mean (μ) = 19 mm
- Standard Deviation (σ) = 5.5 mm
- Maximum allowable size = 23 mm
2. **Calculate the Z-Score:**
The Z-score is a measure of how many standard deviations an element is from the mean. It is calculated as follows:
\[
Z = \frac{X - \mu}{\sigma}
\]
where \( X \) is the maximum allowable size (23 mm in our case).
3. **Substitute the Values:**
\[
Z = \frac{23 - 19}{5.5} = \frac{4}{5.5} \approx 0.727
\]
4. **Use the Z-Score to Find the Probability:**
Using the Z-table or normal distribution calculator, find the cumulative probability corresponding to the Z-score (0.727).
5. **Interpret the Result:**
The cumulative probability represents the proportion of the aggregate size that is within the acceptable range (i.e., less than or equal to 23 mm).
---
### Solution:
After calculating the Z-score:
- Z = 0.727
Looking up the Z-score in the Z-table gives a probability approximately equal to 0.7666.
This means:
- About 76.66% of the aggregate load is usable if we set the maximum size at 23 millimeters.
**Conclusion:**
In this scenario, approximately 76.66% of the recently delivered aggregate load falls within the acceptable size range for road construction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabd5e741-0235-409f-80de-883f5f0e5da7%2Ff351a9e8-130b-44d2-902c-e0dd0455192c%2F3zux5en_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Aggregate Size in Road Construction
---
#### Problem Statement:
The aggregate used in road construction begins to degrade the integrity of the roadway when pieces exceed 23.5 millimeters in diameter. Any pieces smaller than 23.5 millimeters in diameter are considered acceptable.
The recently delivered load of aggregate has a mean size of 19 millimeters and a standard deviation of 5.5 millimeters.
> **Question:**
> What percent of the load is usable if we set the maximum allowable size as 23 millimeters?
---
To determine the usable percentage of the aggregate load, we will need to use the principles of normal distribution.
### Steps to Solve the Problem:
1. **Identify the Given Data:**
- Mean (μ) = 19 mm
- Standard Deviation (σ) = 5.5 mm
- Maximum allowable size = 23 mm
2. **Calculate the Z-Score:**
The Z-score is a measure of how many standard deviations an element is from the mean. It is calculated as follows:
\[
Z = \frac{X - \mu}{\sigma}
\]
where \( X \) is the maximum allowable size (23 mm in our case).
3. **Substitute the Values:**
\[
Z = \frac{23 - 19}{5.5} = \frac{4}{5.5} \approx 0.727
\]
4. **Use the Z-Score to Find the Probability:**
Using the Z-table or normal distribution calculator, find the cumulative probability corresponding to the Z-score (0.727).
5. **Interpret the Result:**
The cumulative probability represents the proportion of the aggregate size that is within the acceptable range (i.e., less than or equal to 23 mm).
---
### Solution:
After calculating the Z-score:
- Z = 0.727
Looking up the Z-score in the Z-table gives a probability approximately equal to 0.7666.
This means:
- About 76.66% of the aggregate load is usable if we set the maximum size at 23 millimeters.
**Conclusion:**
In this scenario, approximately 76.66% of the recently delivered aggregate load falls within the acceptable size range for road construction.
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