Math Problem Statement
Solve this question fully mathematically and carefully.
Solution
To solve this problem, we are given a probability density function (PDF) defined as:
\frac{1}{4} e^{-x/4} & \text{for } x \geq 0 \\ 0 & \text{elsewhere} \end{cases}$$ This describes the length of time $$Y$$ in the air for an airplane to obtain clearance for takeoff, where $$Y = 3X + 2$$. Let's solve the required parts step by step: ### a) Mean length $$E(y)$$ The mean of the random variable $$Y = 3X + 2$$ can be found using the linearity of expectation: $$E(Y) = E(3X + 2) = 3E(X) + 2$$ First, let's find $$E(X)$$. For an exponential distribution $$f(x) = \frac{1}{\lambda} e^{-x/\lambda}$$ where $$\lambda = 4$$: $$E(X) = \lambda = 4$$ So, $$E(Y) = 3(4) + 2 = 12 + 2 = 14$$ ### b) Variance $$\sigma^2(y)$$ The variance of $$Y = 3X + 2$$ is calculated as: $$\text{Var}(Y) = \text{Var}(3X + 2) = 3^2 \text{Var}(X) = 9 \text{Var}(X)$$ For an exponential distribution, the variance is: $$\text{Var}(X) = \lambda^2 = 16$$ Therefore, $$\text{Var}(Y) = 9 \times 16 = 144$$ ### c) Standard Deviation $$\sigma(y)$$ The standard deviation is the square root of the variance: $$\sigma(Y) = \sqrt{144} = 12$$ ### d) Probability $$P(Y \leq 1.5)$$ To find $$P(Y \leq 1.5)$$, we first need to convert this into a probability involving $$X$$: Given $$Y = 3X + 2$$, we solve for $$X$$ when $$Y = 1.5$$: $$1.5 = 3X + 2 \quad \Rightarrow \quad 3X = 1.5 - 2 = -0.5 \quad \Rightarrow \quad X = -\frac{1}{6}$$ Since $$X \geq 0$$ in the PDF, $$P(Y \leq 1.5) = 0$$ because the value of $$X = -\frac{1}{6}$$ is not in the domain of the distribution. So, $$P(Y \leq 1.5) = 0$$ ### Summary of Results: 1. **Mean** $$E(Y) = 14$$ 2. **Variance** $$\text{Var}(Y) = 144$$ 3. **Standard Deviation** $$\sigma(Y) = 12$$ 4. **Probability** $$P(Y \leq 1.5) = 0$$ Do you want more details or have any questions? Here are some related questions: 1. How would the mean change if $$Y = 5X + 3$$? 2. What would be the probability $$P(Y > 20)$$? 3. How would the variance be affected if $$Y = 2X + 1$$? 4. What is the cumulative distribution function (CDF) for $$Y$$? 5. Can you find the mode of $$Y$$ given the exponential distribution of $$X$$? **Tip:** Remember that the linear transformation of a random variable affects both the mean and variance, but only scaling by a factor (not adding constants) affects the variance.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Density Function
Expected Value
Variance
Standard Deviation
Probability Calculations
Formulas
Mean (Expected Value): E(Y) = aE(X) + b
Variance: Var(Y) = a²Var(X)
Standard Deviation: σ(Y) = √Var(Y)
Exponential Distribution Mean: E(X) = λ
Exponential Distribution Variance: Var(X) = λ²
Probability Calculation using transformation
Theorems
Linearity of Expectation
Transformation of Random Variables
Suitable Grade Level
Undergraduate
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