1) The acceptability of a capillary tube for a freezer application is found by measuring the pressure drop in pounds per square inch between the two ends of the tube. The pressures obtained from a manufacturing process of capillary tubes is found to be normally distributed with an average pressure of 130 pounds per square inch and a standard deviation of 4 pounds per square inch. Determine: a) What percent of the pressures are below 121 pounds per square inch? b) What percent of the pressure readings lie between 121 and 134 pounds per square inch?
1) The acceptability of a capillary tube for a freezer application is found by measuring the pressure drop in pounds per square inch between the two ends of the tube. The pressures obtained from a manufacturing process of capillary tubes is found to be normally distributed with an average pressure of 130 pounds per square inch and a standard deviation of 4 pounds per square inch. Determine: a) What percent of the pressures are below 121 pounds per square inch? b) What percent of the pressure readings lie between 121 and 134 pounds per square inch?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Understanding Capillary Tube Acceptability for Freezer Applications**
To determine the acceptability of a capillary tube in a freezer application, we measure the pressure drop in pounds per square inch (psi) between the two ends of the tube. The pressures obtained from a manufacturing process of capillary tubes are found to be normally distributed, with an average pressure of 130 psi and a standard deviation of 4 psi.
**Key Question:**
What percentage of the pressures fall within certain ranges?
**Problem Breakdown:**
**Given Data:**
- Average Pressure (µ): 130 psi
- Standard Deviation (σ): 4 psi
**Sub-Problems:**
**a) What percent of the pressures are below 121 psi?**
To find this percentage, calculate the Z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) = 121 psi
- \( \mu \) = 130 psi
- \( \sigma \) = 4 psi
\[ Z = \frac{121 - 130}{4} = \frac{-9}{4} = -2.25 \]
Using the Z-table, find the cumulative probability corresponding to \( Z = -2.25 \).
**b) What percent of the pressure readings lie between 121 and 134 psi?**
First, calculate the Z-scores for 121 psi and 134 psi.
For 121 psi:
\[ Z_{121} = \frac{121 - 130}{4} = -2.25 \]
For 134 psi:
\[ Z_{134} = \frac{134 - 130}{4} = 1.00 \]
Next, find the cumulative probabilities for these Z-scores using the Z-table. The desired percentage will be the difference between these probabilities.
**Graphical Representation:**
Visualize these problems on a standard normal distribution curve, where:
- The area to the left of \( Z = -2.25 \) represents the percentage of pressures below 121 psi.
- The area between \( Z = -2.25 \) and \( Z = 1.00 \) represents the percentage of pressures between 121 and 134 psi.
These analyses help in understanding the distribution of pressure readings, crucial for ensuring the capillary tubes meet the desired specifications in freezer applications.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7824f8d0-65fb-442c-84f7-7ea5547717e1%2Feb2747c2-e436-40cd-9c87-63382f02a845%2Fc84rkyn_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Capillary Tube Acceptability for Freezer Applications**
To determine the acceptability of a capillary tube in a freezer application, we measure the pressure drop in pounds per square inch (psi) between the two ends of the tube. The pressures obtained from a manufacturing process of capillary tubes are found to be normally distributed, with an average pressure of 130 psi and a standard deviation of 4 psi.
**Key Question:**
What percentage of the pressures fall within certain ranges?
**Problem Breakdown:**
**Given Data:**
- Average Pressure (µ): 130 psi
- Standard Deviation (σ): 4 psi
**Sub-Problems:**
**a) What percent of the pressures are below 121 psi?**
To find this percentage, calculate the Z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) = 121 psi
- \( \mu \) = 130 psi
- \( \sigma \) = 4 psi
\[ Z = \frac{121 - 130}{4} = \frac{-9}{4} = -2.25 \]
Using the Z-table, find the cumulative probability corresponding to \( Z = -2.25 \).
**b) What percent of the pressure readings lie between 121 and 134 psi?**
First, calculate the Z-scores for 121 psi and 134 psi.
For 121 psi:
\[ Z_{121} = \frac{121 - 130}{4} = -2.25 \]
For 134 psi:
\[ Z_{134} = \frac{134 - 130}{4} = 1.00 \]
Next, find the cumulative probabilities for these Z-scores using the Z-table. The desired percentage will be the difference between these probabilities.
**Graphical Representation:**
Visualize these problems on a standard normal distribution curve, where:
- The area to the left of \( Z = -2.25 \) represents the percentage of pressures below 121 psi.
- The area between \( Z = -2.25 \) and \( Z = 1.00 \) represents the percentage of pressures between 121 and 134 psi.
These analyses help in understanding the distribution of pressure readings, crucial for ensuring the capillary tubes meet the desired specifications in freezer applications.
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