If X is a continuous random variable with the following probability 
density function  




 
0 , 2
2 0 2
2
x
a x x , < x
f(x)=
 (i) Find the value of “a”. (ii) Find P[X >1]. 

Given the probability density function (PDF) of the continuous random variable \( X \) as:

\[
f(x) = 
\begin{cases}
a(2 - x^2), & 0 \leq x \leq 2 \\
0, & \text{otherwise}
\end{cases}
\]

### Part (i) - Finding the value of \( a \)

For any probability density function, the total area under the curve must equal 1. That is, the integral of \( f(x) \) over the range of \( X \) must be 1:

\[
\int_{0}^{2} a(2 - x^2) \, dx = 1
\]

Let’s compute this integral:

1. Split the integral:
\[
\int_{0}^{2} a(2 - x^2) \, dx = a \int_{0}^{2} (2 - x^2) \, dx
\]

2. Compute the integral:
\[
a \left[ 2x - \frac{x^3}{3} \right]_0^2 = a \left( 2(2) - \frac{(2)^3}{3} \right)
\]
\[
= a \left( 4 - \frac{8}{3} \right) = a \left( \frac{12}{3} - \frac{8}{3} \right) = a \left( \frac{4}{3} \right)
\]

3. Set this equal to 1:
\[
a \cdot \frac{4}{3} = 1
\]
\[
a = \frac{3}{4}
\]

So, the value of \( a \) is \( \frac{3}{4} \).

### Part (ii) - Finding \( P(X > 1) \)

To find \( P(X > 1) \), we need to compute the integral of \( f(x) \) from 1 to 2:

\[
P(X > 1) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \frac{3}{4} (2 - x^2) \, dx
\]

Let’s compute this integral:

1. Split the integral:
\[
P(X > 1) = \frac{3}{4} \int_{1}^{2} (2 - x^2) \, dx
\]

2. Compute the integral:
\[
\frac{3}{4} \left[ 2x - \frac{x^3}{3} \right]_1^2 = \frac{3}{4} \left( \left( 2(2) - \frac{(2)^3}{3} \right) - \left( 2(1) - \frac{(1)^3}{3} \right) \right)
\]
\[
= \frac{3}{4} \left( \left( 4 - \frac{8}{3} \right) - \left( 2 - \frac{1}{3} \right) \right)
\]
\[
= \frac{3}{4} \left( \left( \frac{12}{3} - \frac{8}{3} \right) - \left( \frac{6}{3} - \frac{1}{3} \right) \right)
\]
\[
= \frac{3}{4} \left( \frac{4}{3} - \frac{5}{3} \right) = \frac{3}{4} \cdot \frac{-1}{3} = -\frac{3}{12} = -\frac{1}{4}
\]

This gives \( P(X > 1) = \frac{1}{4} \). Thus, \( P(X > 1) = 0.25 \).

### Final Answers:
- (i) \( a = \frac{3}{4} \)
- (ii) \( P(X > 1) = 0.25 \)

Would you like more detailed explanations on any step or further questions?

### 5 Follow-Up Questions:
1. What is the significance of normalizing a probability density function?
2. Can you calculate \( P(0.5 \leq X \leq 1.5) \) for the given PDF?
3. How would the PDF change if the range of \( x \) was extended?
4. How can you use this PDF to calculate the expected value of \( X \)?
5. What are the properties of a continuous random variable's PDF?

### Tip:
When integrating polynomials, always double-check your limits and simplify fractions carefully to avoid small calculation errors.

If X is a continuous random variable with the following probability density function: f(x) = a(2 - x^2) for 0 ≤ x ≤ 2 and 0 otherwise, (i) Find the value of 'a'. (ii) Find P[X > 1].

This solution explains how to find the constant 'a' for a given probability density function (PDF) of a continuous random variable and calculates the probability P(X > 1). Key concepts include integration of the PDF and properties of continuous random variables. Suitable for college-level students.

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