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express this using the fonction de repartition

The exercise requires using the cumulative distribution function (CDF), often denoted \( F_X(x) \), to calculate the probabilities for a continuous random variable \( X \) with a given probability density function (PDF). 

The PDF provided for \( X \) is as follows:

\[
f_X(x) = \begin{cases} 
x & \text{if } 0 \leq x \leq 2, \\ 
0 & \text{elsewhere}.
\end{cases}
\]

To solve this using the CDF \( F_X(x) \), let's go through each part:

### Step 1: Find the CDF \( F_X(x) \)

The CDF \( F_X(x) \) is obtained by integrating the PDF from \( -\infty \) up to \( x \):

\[
F_X(x) = \int_{-\infty}^{x} f_X(t) \, dt.
\]

Since \( f_X(x) = 0 \) for \( x < 0 \) and \( x > 2 \), we only need to integrate from 0 to \( x \) within the interval \( 0 \leq x \leq 2 \):

1. **For \( x < 0 \):** \( F_X(x) = 0 \), since \( X \) cannot take values outside \( [0, 2] \).
2. **For \( 0 \leq x \leq 2 \):**
   \[
   F_X(x) = \int_{0}^{x} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{x^2}{2}.
   \]
3. **For \( x \geq 2 \):** \( F_X(x) = 1 \), since the total probability must equal 1.

Thus, the CDF is:

\[
F_X(x) = \begin{cases} 
0 & \text{if } x < 0, \\ 
\frac{x^2}{2} & \text{if } 0 \leq x \leq 2, \\ 
1 & \text{if } x \geq 2.
\end{cases}
\]

### Step 2: Compute the Probabilities Using the CDF

Now, let's solve each part of the question:

#### i) \( P(X > 1) \)

Using the complement rule, \( P(X > 1) = 1 - P(X \leq 1) \):
\[
P(X > 1) = 1 - F_X(1).
\]
Since \( F_X(1) = \frac{1^2}{2} = \frac{1}{2} \),
\[
P(X > 1) = 1 - \frac{1}{2} = \frac{1}{2}.
\]

#### ii) \( P(X < 1) \)

This is simply \( F_X(1) \):
\[
P(X < 1) = F_X(1) = \frac{1}{2}.
\]

#### iii) \( P\left( \frac{1}{2} < X < \frac{3}{2} \right) \)

We calculate this as \( P\left( X < \frac{3}{2} \right) - P\left( X \leq \frac{1}{2} \right) \):
\[
P\left( \frac{1}{2} < X < \frac{3}{2} \right) = F_X\left( \frac{3}{2} \right) - F_X\left( \frac{1}{2} \right).
\]

1. **Calculate \( F_X\left( \frac{3}{2} \right) \):**
   \[
   F_X\left( \frac{3}{2} \right) = \frac{\left( \frac{3}{2} \right)^2}{2} = \frac{\frac{9}{4}}{2} = \frac{9}{8}.
   \]

2. **Calculate \( F_X\left( \frac{1}{2} \right) \):**
   \[
   F_X\left( \frac{1}{2} \right) = \frac{\left( \frac{1}{2} \right)^2}{2} = \frac{\frac{1}{4}}{2} = \frac{1}{8}.
   \]

So,
\[
P\left( \frac{1}{2} < X < \frac{3}{2} \right) = \frac{9}{8} - \frac{1}{8} = \frac{8}{8} = 1.
\]

### Final Answers:

1. \( P(X > 1) = \frac{1}{2} \).
2. \( P(X < 1) = \frac{1}{2} \).
3. \( P\left( \frac{1}{2} < X < \frac{3}{2} \right) = 1 \).

Would you like more details or have any questions?

### Additional Questions:
1. How is the CDF derived from the PDF in other cases?
2. What are some properties of the cumulative distribution function?
3. How does integration help in finding probabilities for continuous random variables?
4. Can the probability of a continuous random variable at a single point be non-zero?
5. How does the complement rule work for calculating probabilities?

### Tip:
Always verify the boundaries of the random variable's support when calculating probabilities to avoid errors.

Express the probabilities in terms of the cumulative distribution function for the given probability density function. Given the PDF: 

\[
f_X(x) = \begin{cases} 
x & \text{if } 0 \leq x \leq 2, \\ 
0 & \text{elsewhere}.
\end{cases}
\]

Find:

1. \( P(X > 1) \)
2. \( P(X < 1) \)
3. \( P\left( \frac{1}{2} < X < \frac{3}{2} \right) \)

Learn how to calculate probabilities for a continuous random variable with a given PDF by using the cumulative distribution function (CDF). This exercise includes calculating P(X > 1), P(X < 1), and P(1/2 < X < 3/2) using integration methods and probability concepts.

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