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The problem asks us to determine three things based on the probability density function (PDF) of a random variable \( X \):

1. **Find \( k \)**
2. **Find \( P(X < 4) \)**
3. **Find \( P(3 < X \leq 6) \)**

The values of \( X \) and the corresponding probabilities are given in the following form:

\[
X: 0, 1, 2, 3, 4, 5, 6
\]
\[
P(X = x): k, 3k, 5k, 7k, 9k, 11k, 13k
\]

### Step 1: Find \( k \)
Since the total probability of a discrete random variable must sum to 1, we can set up the following equation:

\[
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 1
\]

Substituting the probabilities in terms of \( k \):

\[
k + 3k + 5k + 7k + 9k + 11k + 13k = 1
\]

Simplifying the left-hand side:

\[
(1 + 3 + 5 + 7 + 9 + 11 + 13)k = 1
\]

\[
49k = 1
\]

Solving for \( k \):

\[
k = \frac{1}{49}
\]

### Step 2: Find \( P(X < 4) \)
This is the probability that \( X \) takes a value less than 4, i.e., the sum of probabilities for \( X = 0, 1, 2, 3 \):

\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]

Substituting the values in terms of \( k \):

\[
P(X < 4) = k + 3k + 5k + 7k = (1 + 3 + 5 + 7)k = 16k
\]

Now, substitute \( k = \frac{1}{49} \):

\[
P(X < 4) = 16 \times \frac{1}{49} = \frac{16}{49}
\]

### Step 3: Find \( P(3 < X \leq 6) \)
This is the probability that \( X \) is between 3 and 6, inclusive of 6, i.e., \( P(4 \leq X \leq 6) \). This involves summing the probabilities for \( X = 4, 5, 6 \):

\[
P(3 < X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)
\]

Substituting the values in terms of \( k \):

\[
P(3 < X \leq 6) = 9k + 11k + 13k = (9 + 11 + 13)k = 33k
\]

Substituting \( k = \frac{1}{49} \):

\[
P(3 < X \leq 6) = 33 \times \frac{1}{49} = \frac{33}{49}
\]

### Final Answers:
- \( k = \frac{1}{49} \)
- \( P(X < 4) = \frac{16}{49} \)
- \( P(3 < X \leq 6) = \frac{33}{49} \)

---

Would you like further details or explanations on any step? Here are some related questions for further practice:

1. How do you calculate the expected value of \( X \) using this PDF?
2. What is the variance of \( X \) based on the given probabilities?
3. How do cumulative distribution functions (CDF) relate to PDFs in discrete variables?
4. Can you find \( P(X \geq 5) \) from the given PDF?
5. What is the probability that \( X \) equals an odd number?

**Tip:** Always check that the total sum of probabilities equals 1 to confirm that your PDF is valid.

The probability density function of a random variable X is given as: X : 0, 1, 2, 3, 4, 5, 6 P(X = x) : k, 3k, 5k, 7k, 9k, 11k, 13k. Find k, P(X < 4), P(3 < X ≤ 6).

This problem involves finding the constant k and calculating probabilities for a discrete random variable X. Learn how to sum probabilities, apply basic probability theorems, and solve for P(X < 4) and P(3 < X ≤ 6). Suitable for undergraduate students.

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