The net weight in pounds of a packaged chemical herbicide is uniform for 49.68
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![### Uniform Distribution of Packaged Chemical Herbicide Weights
The net weight in pounds of a packaged chemical herbicide is uniformly distributed between 49.68 pounds and 50.24 pounds. Here we explore various statistical aspects of this distribution.
#### (a) Determine the Mean of \( X \)
To find the mean of \( X \) (where X represents the weight), use the formula for the mean of a uniform distribution, which is given by:
\[ \mu = \frac{a + b}{2} \]
- **Input:**
- Lower bound \( a = 49.68 \) pounds
- Upper bound \( b = 50.24 \) pounds
- **Calculation:** Mean \( \mu = \frac{49.68 + 50.24}{2} = 49.96 \) pounds
#### (b) Determine the Variance of \( X \)
The variance of a uniform distribution is given by:
\[ \sigma^2 = \frac{(b - a)^2}{12} \]
- **Input:**
- Lower bound \( a = 49.68 \) pounds
- Upper bound \( b = 50.24 \) pounds
- **Calculation:** Variance \( \sigma^2 = \frac{(50.24 - 49.68)^2}{12} = 0.0317 \) pounds\(^2\)
#### (c) What is \( P(X < 50.1) \)?
To calculate the probability \( P(X < 50.1) \):
Use the cumulative distribution function (CDF) for a uniform distribution:
\[ F(x) = \frac{x - a}{b - a} \]
- **Input:**
- Random variable \( x = 50.1 \) pounds
- Lower bound \( a = 49.68 \)
- Upper bound \( b = 50.24 \)
- **Calculation:**
\( P(X < 50.1) = \frac{50.1 - 49.68}{50.24 - 49.68} \approx 0.750 \)
#### (d) What is \( P(X > 49.9) \)?
To calculate the probability \( P(X > 49.9) \):
Use the property of uniform distribution:
\[ P(X > x) = 1 - F](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F548b861a-42f3-4cfc-8e03-f464d56fea49%2F87a8ff52-8ec8-49d6-bd7c-2bbce5c4b882%2Fam551vm_processed.png&w=3840&q=75)
Transcribed Image Text:### Uniform Distribution of Packaged Chemical Herbicide Weights
The net weight in pounds of a packaged chemical herbicide is uniformly distributed between 49.68 pounds and 50.24 pounds. Here we explore various statistical aspects of this distribution.
#### (a) Determine the Mean of \( X \)
To find the mean of \( X \) (where X represents the weight), use the formula for the mean of a uniform distribution, which is given by:
\[ \mu = \frac{a + b}{2} \]
- **Input:**
- Lower bound \( a = 49.68 \) pounds
- Upper bound \( b = 50.24 \) pounds
- **Calculation:** Mean \( \mu = \frac{49.68 + 50.24}{2} = 49.96 \) pounds
#### (b) Determine the Variance of \( X \)
The variance of a uniform distribution is given by:
\[ \sigma^2 = \frac{(b - a)^2}{12} \]
- **Input:**
- Lower bound \( a = 49.68 \) pounds
- Upper bound \( b = 50.24 \) pounds
- **Calculation:** Variance \( \sigma^2 = \frac{(50.24 - 49.68)^2}{12} = 0.0317 \) pounds\(^2\)
#### (c) What is \( P(X < 50.1) \)?
To calculate the probability \( P(X < 50.1) \):
Use the cumulative distribution function (CDF) for a uniform distribution:
\[ F(x) = \frac{x - a}{b - a} \]
- **Input:**
- Random variable \( x = 50.1 \) pounds
- Lower bound \( a = 49.68 \)
- Upper bound \( b = 50.24 \)
- **Calculation:**
\( P(X < 50.1) = \frac{50.1 - 49.68}{50.24 - 49.68} \approx 0.750 \)
#### (d) What is \( P(X > 49.9) \)?
To calculate the probability \( P(X > 49.9) \):
Use the property of uniform distribution:
\[ P(X > x) = 1 - F
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