The net weight in pounds of a packaged chemical herbicide is uniform for 49.68

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
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Author:Carter
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Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 22PFA
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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
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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