The probability density function of the length of a cutting blade is f(x) = 1.25 for 74.6 < x < 75.4 millimeters. Determine the following: a) P(X < 75.1) = i (Round the answer to 3 decimal places.) b) P(X < 74.8 or X > 75.2) = ₁ c) If the specifications for this process are from 74.7 to 75.3 millimeters, what proportion of cutting blades meets specifications?
The probability density function of the length of a cutting blade is f(x) = 1.25 for 74.6 < x < 75.4 millimeters. Determine the following: a) P(X < 75.1) = i (Round the answer to 3 decimal places.) b) P(X < 74.8 or X > 75.2) = ₁ c) If the specifications for this process are from 74.7 to 75.3 millimeters, what proportion of cutting blades meets specifications?
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![## Probability Density Function of Cutting Blade Length
### Problem Statement
The probability density function of the length of a cutting blade is \( f(x) = 1.25 \) for \( 74.6 < x < 75.4 \) millimeters. Determine the following:
### Questions
**a) \( P(X < 75.1) \) =**
*Round the answer to 3 decimal places.*
**b) \( P(X < 74.8 \text{ or } X > 75.2) \) =**
*Round the answer to 3 decimal places.*
**c)** If the specifications for this process are from 74.7 to 75.3 millimeters, what proportion of cutting blades meets specifications?
---
### Explanation
For those working with probability density functions (PDFs), particularly with uniform distributions, here is a detailed explanation to help you solve the above questions:
1. **Understanding the Given PDF:**
- The function \( f(x) = 1.25 \) is given for the interval \( 74.6 < x < 75.4 \).
- This implies that the length of the cutting blade is uniformly distributed between 74.6 and 75.4 millimeters.
2. **Uniform Distribution Properties:**
- For a uniform distribution over an interval \([a, b]\), the PDF is \( f(x) = \frac{1}{b-a} \).
- In this case, \( b - a = 75.4 - 74.6 = 0.8 \).
- Therefore, \( f(x) = \frac{1}{0.8} = 1.25 \). This matches the given PDF.
3. **Solving the Questions:**
**(a)** \( P(X < 75.1) \)
- To find this probability, integrate the PDF from \( 74.6 \) to \( 75.1 \).
- \( P(X < 75.1) = \int_{74.6}^{75.1} 1.25 \, dx \)
- \( P(X < 75.1) = 1.25 \times (75.1 - 74.6) \)
- \( P(X < 75.1) = 1.25 \times 0.5 \)
- \( P(X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbb0b53a-c5e9-4f61-9041-8945e4ef01cb%2F30b8fd85-cbbb-4d52-87e5-3656a0ebc15a%2F43kb2w9_processed.png&w=3840&q=75)
Transcribed Image Text:## Probability Density Function of Cutting Blade Length
### Problem Statement
The probability density function of the length of a cutting blade is \( f(x) = 1.25 \) for \( 74.6 < x < 75.4 \) millimeters. Determine the following:
### Questions
**a) \( P(X < 75.1) \) =**
*Round the answer to 3 decimal places.*
**b) \( P(X < 74.8 \text{ or } X > 75.2) \) =**
*Round the answer to 3 decimal places.*
**c)** If the specifications for this process are from 74.7 to 75.3 millimeters, what proportion of cutting blades meets specifications?
---
### Explanation
For those working with probability density functions (PDFs), particularly with uniform distributions, here is a detailed explanation to help you solve the above questions:
1. **Understanding the Given PDF:**
- The function \( f(x) = 1.25 \) is given for the interval \( 74.6 < x < 75.4 \).
- This implies that the length of the cutting blade is uniformly distributed between 74.6 and 75.4 millimeters.
2. **Uniform Distribution Properties:**
- For a uniform distribution over an interval \([a, b]\), the PDF is \( f(x) = \frac{1}{b-a} \).
- In this case, \( b - a = 75.4 - 74.6 = 0.8 \).
- Therefore, \( f(x) = \frac{1}{0.8} = 1.25 \). This matches the given PDF.
3. **Solving the Questions:**
**(a)** \( P(X < 75.1) \)
- To find this probability, integrate the PDF from \( 74.6 \) to \( 75.1 \).
- \( P(X < 75.1) = \int_{74.6}^{75.1} 1.25 \, dx \)
- \( P(X < 75.1) = 1.25 \times (75.1 - 74.6) \)
- \( P(X < 75.1) = 1.25 \times 0.5 \)
- \( P(X
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