SOLUTION: Suppose that Y is a continuous random variable whose pdf is given by f (y) = K(4y − 2y^2), 0 < y < 2 0, elsewhere. (a) What is the value of K? (b) Find P(Y

SOLUTION: Suppose that Y is a continuous random variable whose pdf is given by f (y) = K(4y − 2y^2), 0 < y < 2 0, elsewhere. (a) What is the value of K? (b) Find P(Y > 1). (c) F

Question 1177613: Suppose that Y is a continuous random variable whose pdf is given by
f (y) =
K(4y − 2y^2), 0 < y < 2
0, elsewhere.

(a) What is the value of K?
(b) Find P(Y > 1).
(c) Find F (y)
thank you
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Absolutely, let's break down the problem step by step.
**(a) Finding the value of K**
To find the value of K, we'll use the property that the integral of a probability density function (PDF) over its entire range must equal 1.
Therefore, we need to solve the following equation for K:
```
∫[0,2] K(4y - 2y^2) dy = 1
```
Evaluating the integral:
```
K [2y^2 - (2/3)y^3] from 0 to 2 = 1
K (8 - (16/3)) = 1
K (8/3) = 1
```
Solving for K:
```
K = 3/8
```
**(b) Finding P(Y > 1)**
To find P(Y > 1), we'll integrate the PDF from 1 to 2:
```
P(Y > 1) = ∫[1,2] (3/8)(4y - 2y^2) dy
= (3/8) [2y^2 - (2/3)y^3] from 1 to 2
= (3/8) [(8 - (16/3)) - (2 - (2/3))]
= (3/8) (2/3) = 1/4
```
Therefore, P(Y > 1) = 1/4.
**(c) Finding F(y)**
The cumulative distribution function (CDF), F(y), is the integral of the PDF from negative infinity to y. We can define it piecewise:
For y ≤ 0:
```
F(y) = 0
```
For 0 < y < 2:
```
F(y) = ∫[0,y] (3/8)(4t - 2t^2) dt
= (3/8) [2t^2 - (2/3)t^3] from 0 to y
= (3/8) (2y^2 - (2/3)y^3)
= (3/4)y^2 - (1/4)y^3
```
For y ≥ 2:
```
F(y) = 1
```
So, the complete CDF is:
```
F(y) =
0, y ≤ 0
(3/4)y^2 - (1/4)y^3, 0 < y < 2
1, y ≥ 2
```
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