Question 1193031: 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).
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! Let's solve each part step by step.
### Given:
The probability density function \( f(y) = cy \) for \( 0 \leq y \leq 2 \), and \( f(y) = 0 \) elsewhere.
---
### **(a) Find the value of \( c \)**
To be a valid probability density function (PDF), the total probability over the range of \( y \) must equal 1:
\[
\int_{-\infty}^{\infty} f(y) \, dy = 1
\]
Substituting the definition of \( f(y) \):
\[
\int_{0}^{2} cy \, dy = 1
\]
Compute the integral:
\[
\int_{0}^{2} cy \, dy = c \int_{0}^{2} y \, dy = c \left[ \frac{y^2}{2} \right]_0^2 = c \cdot \frac{2^2}{2} = c \cdot 2
\]
Set the result equal to 1:
\[
2c = 1 \quad \implies \quad c = \frac{1}{2}
\]
---
### **(b) Find \( F(y) \), the cumulative distribution function**
The cumulative distribution function \( F(y) \) is the integral of \( f(y) \) from \( -\infty \) to \( y \). For \( 0 \leq y \leq 2 \):
\[
F(y) = \int_{0}^{y} \frac{1}{2}t \, dt
\]
Compute the integral:
\[
F(y) = \frac{1}{2} \int_{0}^{y} t \, dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_0^y = \frac{1}{2} \cdot \frac{y^2}{2} = \frac{y^2}{4}
\]
Thus:
\[
F(y) =
\begin{cases}
0 & \text{if } y < 0 \\
\frac{y^2}{4} & \text{if } 0 \leq y \leq 2 \\
1 & \text{if } y > 2
\end{cases}
\]
---
### **(c) Graph \( f(y) \) and \( F(y) \)**
We will graph \( f(y) = \frac{1}{2}y \) for \( 0 \leq y \leq 2 \) and \( F(y) \) as defined above. Let me generate the graphs for you.
The graphs of \( f(y) \) and \( F(y) \) are displayed:
- \( f(y) \) (Probability Density Function) is a straight line increasing from 0 to 1 at \( y = 2 \).
- \( F(y) \) (Cumulative Distribution Function) is a parabola starting from 0 at \( y = 0 \) and reaching 1 at \( y = 2 \).
---
### **(d) Use \( F(y) \) to find \( P(1 \leq Y \leq 2) \)**
The probability can be calculated as:
\[
P(1 \leq Y \leq 2) = F(2) - F(1)
\]
From \( F(y) = \frac{y^2}{4} \):
\[
F(2) = \frac{2^2}{4} = 1, \quad F(1) = \frac{1^2}{4} = 0.25
\]
Thus:
\[
P(1 \leq Y \leq 2) = 1 - 0.25 = 0.75
\]
---
### **(e) Use \( f(y) \) and geometry to find \( P(1 \leq Y \leq 2) \)**
The probability \( P(1 \leq Y \leq 2) \) can also be found by integrating \( f(y) \) over \( [1, 2] \):
\[
P(1 \leq Y \leq 2) = \int_{1}^{2} \frac{1}{2}y \, dy
\]
Compute the integral:
\[
\int_{1}^{2} \frac{1}{2}y \, dy = \frac{1}{2} \int_{1}^{2} y \, dy = \frac{1}{2} \left[ \frac{y^2}{2} \right]_1^2 = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right)
\]
\[
= \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} = 0.75
\]
---
### Final Answers:
1. \( c = \frac{1}{2} \)
2. \( F(y) = \begin{cases}
0 & y < 0 \\
\frac{y^2}{4} & 0 \leq y \leq 2 \\
1 & y > 2
\end{cases} \)
3. Graphs are shown above.
4. \( P(1 \leq Y \leq 2) = 0.75 \) (using \( F(y) \)).
5. \( P(1 \leq Y \leq 2) = 0.75 \) (using \( f(y) \) and geometry).
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