The following table represents the probability of the number of cars owned by a college student. Give your answer to at least 3 decimal places.
x 	P(x)
0 	0.48
1 	0.35
2 	0.13
3 	0.04

a) Is this a valid discrete probability distribution?

b) Find the mean number of cars owned.

0.73 

c) Find the standard deviation of the number of cars owned.


Let's analyze each part of the problem step by step.  

### **Given:**  
- \(x\): Number of cars owned by a college student  
- \(P(x)\): Probability associated with each number of cars  
\[
\begin{aligned}
x & = 0, 1, 2, 3 \\
P(x) & = 0.48, 0.35, 0.13, 0.04
\end{aligned}
\]

---

### **a) Is this a valid discrete probability distribution?**

A probability distribution is valid if:
1. Each probability \(P(x)\) is between 0 and 1.  
2. The sum of all probabilities equals 1.

Check the sum of probabilities:  
\[
0.48 + 0.35 + 0.13 + 0.04 = 1.00
\]

Since all probabilities are between 0 and 1, and the total equals 1, **this is a valid discrete probability distribution.**

---

### **b) Find the mean number of cars owned.**

The mean \(\mu\) of a discrete probability distribution is calculated using the formula:  
\[
\mu = \sum x \cdot P(x)
\]

Plug in the values:  
\[
\mu = (0 \times 0.48) + (1 \times 0.35) + (2 \times 0.13) + (3 \times 0.04)
\]
\[
\mu = 0 + 0.35 + 0.26 + 0.12 = 0.73
\]

**Mean \(\mu = 0.73\) cars.** ✅

---

### **c) Find the standard deviation of the number of cars owned.**

The standard deviation \(\sigma\) is calculated using:  
\[
\sigma = \sqrt{\sum [x^2 \cdot P(x)] - \mu^2}
\]

First, compute \(x^2 \cdot P(x)\) for each value:  
\[
0^2 \times 0.48 = 0, \quad 1^2 \times 0.35 = 0.35, \quad 2^2 \times 0.13 = 0.52, \quad 3^2 \times 0.04 = 0.36
\]

Sum these values:  
\[
0 + 0.35 + 0.52 + 0.36 = 1.23
\]

Now, use the standard deviation formula:  
\[
\sigma = \sqrt{1.23 - (0.73)^2}
\]
\[
\sigma = \sqrt{1.23 - 0.5329}
\]
\[
\sigma = \sqrt{0.6971}
\]
\[
\sigma \approx 0.835
\]

**Standard deviation \(\sigma \approx 0.835\) cars.** ✅

---

### ✅ **Final Answer:**  
a) **Yes**, it's a valid discrete probability distribution.  
b) **Mean:** \(\mu = 0.73\) cars.  
c) **Standard deviation:** \(\sigma = 0.835\) cars.

---

💡 **Would you like a more detailed breakdown of the formulas or additional examples for practice?**  

---

### 💡 **Follow-Up Questions:**  
1. How would the mean change if the probability of owning zero cars decreased?  
2. Can the standard deviation ever be greater than the mean in a discrete distribution?  
3. What does the standard deviation tell us about the number of cars owned?  
4. How would adding another category (e.g., 4 cars) affect the distribution?  
5. Could this probability distribution be represented graphically using a histogram?  

**Tip:** When calculating standard deviation, always square and sum before subtracting the square of the mean. This avoids negative values inside the square root.

The following table represents the probability of the number of cars owned by a college student. Give your answer to at least 3 decimal places.

x  P(x)
0  0.48
1  0.35
2  0.13
3  0.04

a) Is this a valid discrete probability distribution?
b) Find the mean number of cars owned.

0.73

c) Find the standard deviation of the number of cars owned.

Learn how to verify if a probability distribution is valid and calculate the mean and standard deviation for the number of cars owned by college students. Includes a step-by-step breakdown of calculations and relevant formulas for expected value and standard deviation. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation