The p.d.f of a continuous random variable X is given by f(x) = {

x+1

k

, 2 ≤ x ≤ 4

0 otherwise

.

(a) Find k. (b) Obtain the distribution function F(x) (c) Using F(x) compute P( X ≥ 3)

(d) Find Mean and Variance of X (e) Using F(x) find P (2.5 < X < 3).

The given probability density function (p.d.f) \( f(x) \) is described as follows:

\[
f(x) = 
\begin{cases}
\frac{x+1}{k}, & 2 \leq x \leq 4 \\
0, & \text{otherwise}
\end{cases}
\]

We'll go step by step to solve each part:

### (a) Find \( k \)

For \( f(x) \) to be a valid probability density function, the total area under the curve must be 1. This means:

\[
\int_{-\infty}^{\infty} f(x) \, dx = 1
\]

Given that \( f(x) = 0 \) outside the interval \( 2 \leq x \leq 4 \), the equation reduces to:

\[
\int_{2}^{4} \frac{x+1}{k} \, dx = 1
\]

Let's solve for \( k \):

\[
\frac{1}{k} \int_{2}^{4} (x+1) \, dx = 1
\]

First, compute the integral:

\[
\int_{2}^{4} (x+1) \, dx = \left[ \frac{x^2}{2} + x \right]_{2}^{4} = \left( \frac{4^2}{2} + 4 \right) - \left( \frac{2^2}{2} + 2 \right)
\]
\[
= \left( \frac{16}{2} + 4 \right) - \left( \frac{4}{2} + 2 \right) = (8 + 4) - (2 + 2) = 12 - 4 = 8
\]

Thus, we have:

\[
\frac{1}{k} \times 8 = 1 \quad \Rightarrow \quad k = 8
\]

### (b) Obtain the distribution function \( F(x) \)

The cumulative distribution function (C.D.F) \( F(x) \) is found by integrating the p.d.f. \( f(x) \) from the lower bound of the distribution (here, 2) to \( x \).

\[
F(x) = \int_{2}^{x} f(t) \, dt = \int_{2}^{x} \frac{t+1}{8} \, dt
\]

For \( 2 \leq x \leq 4 \), we can compute the integral:

\[
F(x) = \frac{1}{8} \int_{2}^{x} (t+1) \, dt = \frac{1}{8} \left[ \frac{t^2}{2} + t \right]_{2}^{x}
\]

Substitute the limits:

\[
F(x) = \frac{1}{8} \left( \left( \frac{x^2}{2} + x \right) - \left( \frac{2^2}{2} + 2 \right) \right)
\]
\[
= \frac{1}{8} \left( \frac{x^2}{2} + x - (2 + 2) \right) = \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right)
\]

Thus, the C.D.F. for \( 2 \leq x \leq 4 \) is:

\[
F(x) = \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right)
\]

For \( x < 2 \), \( F(x) = 0 \), and for \( x > 4 \), \( F(x) = 1 \).

### (c) Using \( F(x) \), compute \( P(X \geq 3) \)

Using the C.D.F., the probability \( P(X \geq 3) \) is:

\[
P(X \geq 3) = 1 - F(3)
\]

Let's compute \( F(3) \) using the expression for \( F(x) \):

\[
F(3) = \frac{1}{8} \left( \frac{3^2}{2} + 3 - 4 \right) = \frac{1}{8} \left( \frac{9}{2} + 3 - 4 \right)
\]
\[
= \frac{1}{8} \left( 4.5 + 3 - 4 \right) = \frac{1}{8} \times 3.5 = \frac{3.5}{8}
\]

Thus:

\[
F(3) = 0.4375
\]

Now, calculate \( P(X \geq 3) \):

\[
P(X \geq 3) = 1 - 0.4375 = 0.5625
\]

### (d) Find Mean and Variance of \( X \)

The mean \( \mu \) of a continuous random variable is given by:

\[


The p.d.f of a continuous random variable X is given by f(x) = { (x+1)/k, 2 ≤ x ≤ 4; 0 otherwise. (a) Find k. (b) Obtain the distribution function F(x) (c) Using F(x), compute P(X ≥ 3). (d) Find the Mean and Variance of X. (e) Using F(x), find P(2.5 < X < 3).

This problem explores the p.d.f of a continuous random variable X, covering the steps to find the constant k, the cumulative distribution function (CDF), and probabilities such as P(X ≥ 3). Learn how to calculate mean and variance using integral calculus.

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