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
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:**
\[ P(9.4 < X < 11.2) = \int_{9.4}^{11.2} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7167d01a-fc0a-4c3f-ab49-29dc29956a80%2F7e6ffa73-86f8-49a4-bf28-0b96df728a70%2Fsas6l6n_processed.png&w=3840&q=75)
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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