Find the proportion X of individuals who can be expected to respond to a certain mail-order solicitation if X has the density function below. f(x) = 2(x+4) 9 0₁ 0

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**Title: Understanding Probability Density Functions in Mail-Order Solicitations**

**Introduction:**
Learn how to calculate the expected proportion of individuals who respond to a mail-order solicitation using probability density functions. 

**Problem Statement:**
Find the proportion \(X\) of individuals expected to respond to a certain mail-order solicitation if \(X\) has the density function defined below.

**Density Function:**

\[ f(x) = \begin{cases} 
\frac{2(x+4)}{9}, & 0 < x < 1, \\
0, & \text{elsewhere.}
\end{cases} \]

**Explanation:**
- **Range of \(x\):** The function is defined for \(x\) between 0 and 1, representing possible values of the proportion \(X\) that respond.
- **Elsewhere Condition:** The probability density is 0 for values of \(x\) outside the interval (0,1).

**Objective:**
Calculate the proportion of individuals responding using this defined density function. Integrate \(f(x)\) over its range to find the probability.
Transcribed Image Text:**Title: Understanding Probability Density Functions in Mail-Order Solicitations** **Introduction:** Learn how to calculate the expected proportion of individuals who respond to a mail-order solicitation using probability density functions. **Problem Statement:** Find the proportion \(X\) of individuals expected to respond to a certain mail-order solicitation if \(X\) has the density function defined below. **Density Function:** \[ f(x) = \begin{cases} \frac{2(x+4)}{9}, & 0 < x < 1, \\ 0, & \text{elsewhere.} \end{cases} \] **Explanation:** - **Range of \(x\):** The function is defined for \(x\) between 0 and 1, representing possible values of the proportion \(X\) that respond. - **Elsewhere Condition:** The probability density is 0 for values of \(x\) outside the interval (0,1). **Objective:** Calculate the proportion of individuals responding using this defined density function. Integrate \(f(x)\) over its range to find the probability.
### Problem Statement:

If a dealer's profit, in units of $3000, on a new automobile can be viewed as a random variable \( X \) with the density function given below, find the average profit per automobile.

### Density Function:

\[ 
f(x) = \begin{cases} 
\frac{1}{12}(7 - x), & 0 < x < 2, \\
0, & \text{elsewhere}
\end{cases} 
\]

### Explanation:

The function \( f(x) \) is defined as follows:
- Within the interval \( 0 < x < 2 \), the probability density function is \( \frac{1}{12}(7 - x) \).
- For values of \( x \) outside this interval, the density function is 0, indicating that profit values outside this range are not considered in this model.

To find the average (or expected) profit per automobile, you need to calculate the expected value \( E(X) \) of the random variable \( X \) using the density function provided. This involves integrating the product of \( x \) and \( f(x) \) over the interval from 0 to 2:

\[ 
E(X) = \int_0^2 x \cdot f(x) \, dx 
\]

The expected value will give the average profit in units of $3000. Multiply the result by $3000 to find the average profit in dollars.
Transcribed Image Text:### Problem Statement: If a dealer's profit, in units of $3000, on a new automobile can be viewed as a random variable \( X \) with the density function given below, find the average profit per automobile. ### Density Function: \[ f(x) = \begin{cases} \frac{1}{12}(7 - x), & 0 < x < 2, \\ 0, & \text{elsewhere} \end{cases} \] ### Explanation: The function \( f(x) \) is defined as follows: - Within the interval \( 0 < x < 2 \), the probability density function is \( \frac{1}{12}(7 - x) \). - For values of \( x \) outside this interval, the density function is 0, indicating that profit values outside this range are not considered in this model. To find the average (or expected) profit per automobile, you need to calculate the expected value \( E(X) \) of the random variable \( X \) using the density function provided. This involves integrating the product of \( x \) and \( f(x) \) over the interval from 0 to 2: \[ E(X) = \int_0^2 x \cdot f(x) \, dx \] The expected value will give the average profit in units of $3000. Multiply the result by $3000 to find the average profit in dollars.
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