If X has a uniform distribution for x=1,2,...,n and V(X)=24, find P(4<X<15).

We are given that \( X \) has a uniform distribution over \( \{ 1, 2, \dots, n \} \), and the variance of \( X \), denoted \( V(X) \), is 24. The goal is to find the probability \( P(4 < X < 15) \).

### Step 1: Mean and Variance of a Uniform Distribution
For a uniform distribution on the integers \( \{ 1, 2, \dots, n \} \), the mean \( E(X) \) and the variance \( V(X) \) are given by:

\[
E(X) = \frac{n + 1}{2}
\]
\[
V(X) = \frac{(n^2 - 1)}{12}
\]

Since we are given that \( V(X) = 24 \), we can set up the equation:

\[
\frac{n^2 - 1}{12} = 24
\]

### Step 2: Solve for \( n \)
Multiply both sides of the equation by 12 to eliminate the denominator:

\[
n^2 - 1 = 288
\]

Add 1 to both sides:

\[
n^2 = 289
\]

Now, take the square root of both sides:

\[
n = \sqrt{289} = 17
\]

So, \( n = 17 \).

### Step 3: Find \( P(4 < X < 15) \)
Since \( X \) is uniformly distributed over \( \{ 1, 2, \dots, 17 \} \), the probability mass function is:

\[
P(X = k) = \frac{1}{17} \quad \text{for each} \quad k \in \{ 1, 2, \dots, 17 \}
\]

We are asked to find \( P(4 < X < 15) \), which means we want the probability that \( X \) takes on a value between 5 and 14 inclusive. Therefore:

\[
P(4 < X < 15) = P(X = 5) + P(X = 6) + \dots + P(X = 14)
\]

The number of integers between 5 and 14 inclusive is \( 14 - 5 + 1 = 10 \).

Thus:

\[
P(4 < X < 15) = \frac{10}{17}
\]

### Final Answer:
\[
P(4 < X < 15) = \frac{10}{17}
\]

Would you like more details or have any questions about the solution?

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Here are 5 related questions to extend this concept:

1. How would you calculate \( P( X \leq 10) \) for this uniform distribution?
2. What is the cumulative distribution function (CDF) for a uniform distribution on integers?
3. How does the variance formula change for a uniform distribution on \( [0, n] \) instead of integers?
4. What is the standard deviation of \( X \) in this problem?
5. How would you calculate \( P( X = 7) \) for this specific distribution?

**Tip:** In uniform distributions, probabilities for intervals can be computed by counting the number of valid outcomes within that interval and dividing by the total number of possible outcomes.

Learn how to solve a uniform distribution problem where V(X) = 24 and find the probability P(4 < X < 15). This step-by-step solution explores key statistical concepts such as variance, probability, and uniform distribution, making it suitable for undergraduate-level students.

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