Math Problem Statement
Suppose that Y possesses the density function f(y) = { cy, 0 ≤ y ≤ 2; 0, elsewhere }.
(a) Find the value of c that makes f(y) a probability density function.
(b) Find F(y).
(c) Graph f(y) and F(y).
(d) Use F(y) to find P(1 ≤ Y ≤ 2).
(e) Use f(y) and geometry to find P(1 ≤ Y ≤ 2).
Solution
Let's break down the problem step by step. The probability density function is defined as:
\begin{cases} cy, & 0 \leq y \leq 2, \\ 0, & \text{elsewhere}. \end{cases}$$ ### Part (a) - Finding $$c$$ For $$f(y)$$ to be a valid probability density function, the total area under the curve of the PDF must equal 1. This means: $$\int_{-\infty}^{\infty} f(y) \, dy = 1.$$ Since $$f(y) = cy$$ from 0 to 2 and 0 elsewhere, we need to integrate $$f(y)$$ from 0 to 2: $$\int_0^2 cy \, dy = 1.$$ Solve this integral: $$\int_0^2 cy \, dy = c \left[\frac{y^2}{2}\right]_0^2 = c \cdot \frac{4}{2} = 2c.$$ Set this equal to 1: $$2c = 1 \implies c = \frac{1}{2}.$$ Thus, the value of $$c$$ is $$\frac{1}{2}$$. ### Part (b) - Finding $$F(y)$$ The cumulative distribution function (CDF), $$F(y)$$, is found by integrating $$f(y)$$ from the lower bound of $$y$$ (in this case, 0) to any given value $$y$$: $$F(y) = \int_0^y f(t) \, dt.$$ For $$0 \leq y \leq 2$$, $$f(t) = \frac{1}{2}t$$, so we integrate: $$F(y) = \int_0^y \frac{1}{2}t \, dt = \frac{1}{2} \left[\frac{t^2}{2}\right]_0^y = \frac{1}{4}y^2.$$ Thus, for $$0 \leq y \leq 2$$, we have: $$F(y) = \frac{1}{4}y^2.$$ For $$y > 2$$, $$F(y) = 1$$ because all the probability is accumulated, and for $$y < 0$$, $$F(y) = 0$$. ### Part (c) - Graphing $$f(y)$$ and $$F(y)$$ - $$f(y) = \frac{1}{2}y$$ is a straight line that starts at $$(0, 0)$$ and ends at $$(2, 1)$$. - $$F(y) = \frac{1}{4}y^2$$ is a parabola starting at $$(0, 0)$$ and ending at $$(2, 1)$$. ### Part (d) - Finding $$P(1 \leq Y \leq 2)$$ using $$F(y)$$ The probability $$P(1 \leq Y \leq 2)$$ is: $$P(1 \leq Y \leq 2) = F(2) - F(1).$$ We know that $$F(2) = 1$$ and: $$F(1) = \frac{1}{4}(1^2) = \frac{1}{4}.$$ Thus: $$P(1 \leq Y \leq 2) = 1 - \frac{1}{4} = \frac{3}{4}.$$ ### Part (e) - Using $$f(y)$$ and geometry to find $$P(1 \leq Y \leq 2)$$ We can calculate $$P(1 \leq Y \leq 2)$$ geometrically by finding the area under the curve $$f(y)$$ from 1 to 2. The function $$f(y) = \frac{1}{2}y$$ is linear, and the area from 1 to 2 forms a trapezoid. The area of a trapezoid is: $$\text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h,$$ where $$b_1 = f(1) = \frac{1}{2} \times 1 = \frac{1}{2}$$, $$b_2 = f(2) = 1$$, and the height $$h = 2 - 1 = 1$$. Thus: $$\text{Area} = \frac{1}{2} \times \left(\frac{1}{2} + 1\right) \times 1 = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}.$$ This confirms the result from part (d): $$P(1 \leq Y \leq 2) = \frac{3}{4}$$. --- Would you like further details or have any questions? Here are five related questions for further exploration: 1. What is the significance of the constant $$c$$ in probability density functions? 2. How would the function change if the interval for $$y$$ were different? 3. Can you calculate the expected value $$E[Y]$$ for this distribution? 4. How would you find the variance of $$Y$$ based on $$f(y)$$? 5. What if the density function was quadratic instead of linear? **Tip:** When dealing with probability density functions, always check that the integral over the possible range equals 1.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Density Functions
Integration
Cumulative Distribution Functions
Area under Curve
Formulas
∫ f(y) dy = 1 (PDF normalization condition)
F(y) = ∫ f(t) dt from 0 to y (CDF definition)
P(a ≤ Y ≤ b) = F(b) - F(a)
Area of Trapezoid = 1/2 × (b1 + b2) × h
Theorems
Fundamental Theorem of Calculus
Probability Density Properties
Geometry of Trapezoids
Suitable Grade Level
Undergraduate / College Level
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