The daily amount of coffee, in liters, dispensed by a machine located in an airport lobby is a random variable X having a continuous uniform distribution with A = 9 and B = 12. Find the probability that on a given day the amount of coffee dispensed by this machine will be (a) at most 11.4 liters; (b) more than 9.4 liters but less than 11.2 liters; (c) at least 10.8 liters. (a) The probability is (Simplify your answer.) (b) The probability is (Simplify your answer.) (c) The probability is (Simplify your answer.) C

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**Title: Probability Calculations with Continuous Uniform Distribution**

**Problem Description:**

The daily amount of coffee, in liters, dispensed by a machine located in an airport lobby is a random variable \(X\) having a continuous uniform distribution with \(A = 9\) and \(B = 12\). 

**Calculate the following probabilities:**
1. The probability that on a given day the amount of coffee dispensed by this machine will be:
   - (a) at most 11.4 liters;
   - (b) more than 9.4 liters but less than 11.2 liters;
   - (c) at least 10.8 liters.

**Detailed Questions and Answers:**

Based on the given continuous uniform distribution properties, we need to solve for the following probabilities with appropriate simplification of the answers:

**(a) The probability is [___].**
*(Simplify your answer.)*

**(b) The probability is [___].**
*(Simplify your answer.)*

**(c) The probability is [___].**
*(Simplify your answer.)*

**Solution Explanation:**

For a continuous uniform distribution \(X \sim \text{Uniform}(A, B)\), the probability density function \(f(x)\) is defined as:

\[ f(x) = \frac{1}{B - A}, \text{ for } A \le x \le B \]

Given \(A = 9\) and \(B = 12\):

\[ f(x) = \frac{1}{12 - 9} = \frac{1}{3} \]

**(a) Probability that \(X\) is at most 11.4 liters:**

\[ P(X \leq 11.4) = \int_{9}^{11.4} f(x) \, dx = \int_{9}^{11.4} \frac{1}{3} \, dx = \frac{11.4 - 9}{3} = \frac{2.4}{3} = 0.8 \]

So, **(a) The probability is \(0.8\).**

**(b) Probability that \(X\) is more than 9.4 liters but less than 11.2 liters:**

\[ P(9.4 < X < 11.2) = \int_{9.4}^{11.2} \
Transcribed Image Text:**Title: Probability Calculations with Continuous Uniform Distribution** **Problem Description:** The daily amount of coffee, in liters, dispensed by a machine located in an airport lobby is a random variable \(X\) having a continuous uniform distribution with \(A = 9\) and \(B = 12\). **Calculate the following probabilities:** 1. The probability that on a given day the amount of coffee dispensed by this machine will be: - (a) at most 11.4 liters; - (b) more than 9.4 liters but less than 11.2 liters; - (c) at least 10.8 liters. **Detailed Questions and Answers:** Based on the given continuous uniform distribution properties, we need to solve for the following probabilities with appropriate simplification of the answers: **(a) The probability is [___].** *(Simplify your answer.)* **(b) The probability is [___].** *(Simplify your answer.)* **(c) The probability is [___].** *(Simplify your answer.)* **Solution Explanation:** For a continuous uniform distribution \(X \sim \text{Uniform}(A, B)\), the probability density function \(f(x)\) is defined as: \[ f(x) = \frac{1}{B - A}, \text{ for } A \le x \le B \] Given \(A = 9\) and \(B = 12\): \[ f(x) = \frac{1}{12 - 9} = \frac{1}{3} \] **(a) Probability that \(X\) is at most 11.4 liters:** \[ P(X \leq 11.4) = \int_{9}^{11.4} f(x) \, dx = \int_{9}^{11.4} \frac{1}{3} \, dx = \frac{11.4 - 9}{3} = \frac{2.4}{3} = 0.8 \] So, **(a) The probability is \(0.8\).** **(b) Probability that \(X\) is more than 9.4 liters but less than 11.2 liters:** \[ P(9.4 < X < 11.2) = \int_{9.4}^{11.2} \
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