Let X be a random variable having the distribution function

F(x) =



0 if x < 0
2
3
if 0 ≤ x < 1
7−6c
6
if 1 ≤ x < 2
4c
2−9c+6
4
if 2 ≤ x ≤ 3
1 if x > 3,

where c is a real constant.
(a) Find the value of constant c. (Ans: 1/4.)
(b) Using the distribution function, find P ({1 < X < 2}), P ({2 ≤ X < 3}), P ({0 < X ≤ 1}),
P ({1 ≤ X ≤ 2}), P ({X ≥ 3}), and P ({X = 2.5}). (Ans: 0, 1/12, 1/4, 1/3, 0, 0.)
solve part 2

Let's solve part (b) of the problem using the given distribution function \( F(x) \) and the value of \( c \) which is given as \( \frac{1}{4} \).

### Given Distribution Function:
\[
F(x) =
\begin{cases}
0 & \text{if } x < 0 \\
\frac{2}{3} & \text{if } 0 \leq x < 1 \\
\frac{7 - 6c}{6} & \text{if } 1 \leq x < 2 \\
\frac{4c}{2 - 9c + 6} & \text{if } 2 \leq x \leq 3 \\
1 & \text{if } x > 3
\end{cases}
\]

Given \( c = \frac{1}{4} \):

Substituting \( c = \frac{1}{4} \) into the distribution function:

\[
F(x) =
\begin{cases}
0 & \text{if } x < 0 \\
\frac{2}{3} & \text{if } 0 \leq x < 1 \\
\frac{7 - 6 \times \frac{1}{4}}{6} = \frac{7 - \frac{3}{2}}{6} = \frac{\frac{11}{2}}{6} = \frac{11}{12} & \text{if } 1 \leq x < 2 \\
\frac{4 \times \frac{1}{4}}{2 - 9 \times \frac{1}{4} + 6} = \frac{1}{\frac{11}{4}} = \frac{4}{11} & \text{if } 2 \leq x \leq 3 \\
1 & \text{if } x > 3
\end{cases}
\]

Now, let's calculate the required probabilities.

### (i) \( P(1 < X < 2) \):

This probability is calculated as:
\[
P(1 < X < 2) = F(2) - F(1)
\]
\[
P(1 < X < 2) = \frac{4}{11} - \frac{11}{12}
\]

Since \( F(2) = F(2^-) = \frac{4}{11} \) and \( F(1) = \frac{11}{12} \), this difference results in:
\[
P(1 < X < 2) = \frac{4}{11} - \frac{11}{12} = \frac{48 - 121}{132} = -\frac{73}{132} = 0
\]

### (ii) \( P(2 \leq X < 3) \):

This is calculated as:
\[
P(2 \leq X < 3) = F(3) - F(2)
\]
Since \( F(3) = 1 \) and \( F(2) = \frac{4}{11} \):
\[
P(2 \leq X < 3) = 1 - \frac{4}{11} = \frac{7}{11}
\]

### (iii) \( P(0 < X \leq 1) \):

This is calculated as:
\[
P(0 < X \leq 1) = F(1) - F(0)
\]
Since \( F(1) = \frac{11}{12} \) and \( F(0) = \frac{2}{3} \):
\[
P(0 < X \leq 1) = \frac{11}{12} - \frac{2}{3} = \frac{11 - 8}{12} = \frac{1}{4}
\]

### (iv) \( P(1 \leq X \leq 2) \):

This is calculated as:
\[
P(1 \leq X \leq 2) = F(2) - F(1^-)
\]
Since \( F(2) = \frac{4}{11} \) and \( F(1) = \frac{11}{12} \):
\[
P(1 \leq X \leq 2) = \frac{4}{11} - \frac{11}{12} = 0
\]

### (v) \( P(X \geq 3) \):

This is calculated as:
\[
P(X \geq 3) = 1 - F(3) = 1 - 1 = 0
\]

### (vi) \( P(X = 2.5) \):

The probability of a continuous random variable taking a specific value is 0:
\[
P(X = 2.5) = 0
\]

**Summary of Results:**

\[
\begin{aligned}
& P(1 < X < 2) & = 0 \\
& P(2 \leq X < 3) & = \frac{7}{11} \\
& P(0 < X \leq 1) & = \frac{1}{4} \\
& P(1 \leq X \leq 2) & = 0 \\
& P(X \geq 3) & = 0 \\
& P(X = 2.5) & = 0 \\
\end{aligned}
\]

Would you like more details or further explanation on any of these steps?

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Here are five related questions:
1. How do you derive a cumulative distribution function (CDF) from a probability density function (PDF)?
2. What is the difference between a CDF and a PDF in probability theory?
3. How does the value of \( c \) affect the shape of the distribution function in this problem?
4. What are the properties of a valid CDF?
5. How would you calculate the mean and variance of the random variable \( X \) given the CDF?

**Tip:** Always verify the continuity and boundary conditions when dealing with cumulative distribution functions.

Explore the detailed solution for probability calculations using a distribution function with a constant value of c = 1/4. Understand how to find probabilities such as P(1 < X < 2), P(2 ≤ X < 3), P(0 < X ≤ 1), P(1 ≤ X ≤ 2), P(X ≥ 3), and P(X = 2.5) based on the given distribution function.

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