Let a continuous random variable X denote the diameter of a hole drilled in a sheet of metal component. The target diameter is 12.5 mm. Most random disturbances to the process result in larger diameters. Historical data show that the distribution of X can be modeled by a pdf f(x)=20e-20(x-12.5), for x≥12.5. If a component with a diameter greater than 12.60 mm is scrapped. a) What portion of parts is scrapped? b) What portion of parts is between 12.5 and 12.6 mm? c) What is the mean and variance of X?
Let a continuous random variable X denote the diameter of a hole drilled in a sheet of metal component. The target diameter is 12.5 mm. Most random disturbances to the process result in larger diameters. Historical data show that the distribution of X can be modeled by a pdf f(x)=20e-20(x-12.5), for x≥12.5. If a component with a diameter greater than 12.60 mm is scrapped. a) What portion of parts is scrapped? b) What portion of parts is between 12.5 and 12.6 mm? c) What is the mean and variance of X?
MATLAB: An Introduction with Applications
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![### Probability Distribution of Hole Diameters in Metal Sheets
Let a continuous random variable \( X \) denote the diameter of a hole drilled in a sheet of metal component. The target diameter is 12.5 mm. Due to random disturbances in the process, the resulting diameters are often larger. Historical data shows that the distribution of \( X \) can be modeled by a probability density function (pdf):
\[ f(x) = 20e^{-20(x-12.5)}, \quad \text{for} \, x \geq 12.5. \]
If a component with a diameter greater than 12.60 mm is scrapped, the following questions need to be answered:
**a) What portion of parts is scrapped?**
**b) What portion of parts has diameters between 12.5 and 12.6 mm?**
**c) What is the mean and variance of \( X \)?**
### Solutions:
**a) Portion of parts scrapped:**
To determine the portion of parts scrapped (with \( x > 12.60 \) mm), we need to calculate the cumulative distribution function (CDF) from 12.5 to 12.6 mm and subtract it from 1.
\[ P(X > 12.6) = 1 - P(X \leq 12.6) \]
Since the CDF function \( F(x) \) is given by the integral of the pdf \( f(x) \):
\[ P(X \leq 12.6) = \int_{12.5}^{12.6} 20e^{-20(x-12.5)} \, dx \]
Let \( u = x - 12.5 \), then:
\[ P(X \leq 12.6) = \int_{0}^{0.1} 20e^{-20u} \, du \]
Evaluating the integral:
\[ P(X \leq 12.6) = \left[ -e^{-20u} \right]_0^{0.1} = -e^{-2} + e^0 = 1 - e^{-2} \approx 1 - 0.1353 = 0.8647 \]
Thus, the portion of parts scrapped is:
\[ P(X > 12.6) = 1 - 0.8647 = 0.1353 \]
**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb19fb600-654f-4bb5-880f-15449a731741%2F974447d1-91dc-4d65-a81e-ade75010e4a4%2F9fg9fzc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Distribution of Hole Diameters in Metal Sheets
Let a continuous random variable \( X \) denote the diameter of a hole drilled in a sheet of metal component. The target diameter is 12.5 mm. Due to random disturbances in the process, the resulting diameters are often larger. Historical data shows that the distribution of \( X \) can be modeled by a probability density function (pdf):
\[ f(x) = 20e^{-20(x-12.5)}, \quad \text{for} \, x \geq 12.5. \]
If a component with a diameter greater than 12.60 mm is scrapped, the following questions need to be answered:
**a) What portion of parts is scrapped?**
**b) What portion of parts has diameters between 12.5 and 12.6 mm?**
**c) What is the mean and variance of \( X \)?**
### Solutions:
**a) Portion of parts scrapped:**
To determine the portion of parts scrapped (with \( x > 12.60 \) mm), we need to calculate the cumulative distribution function (CDF) from 12.5 to 12.6 mm and subtract it from 1.
\[ P(X > 12.6) = 1 - P(X \leq 12.6) \]
Since the CDF function \( F(x) \) is given by the integral of the pdf \( f(x) \):
\[ P(X \leq 12.6) = \int_{12.5}^{12.6} 20e^{-20(x-12.5)} \, dx \]
Let \( u = x - 12.5 \), then:
\[ P(X \leq 12.6) = \int_{0}^{0.1} 20e^{-20u} \, du \]
Evaluating the integral:
\[ P(X \leq 12.6) = \left[ -e^{-20u} \right]_0^{0.1} = -e^{-2} + e^0 = 1 - e^{-2} \approx 1 - 0.1353 = 0.8647 \]
Thus, the portion of parts scrapped is:
\[ P(X > 12.6) = 1 - 0.8647 = 0.1353 \]
**
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