The probability distribution for the number of sales for a salesman each day is given below. What is the expected number of sales for this salesman? Round to the nearest tenth. n sales Probability of n sales Paragraph V 5 0.40 BI U A/ h 7 0.35 V + v 9 0.25 ...
The probability distribution for the number of sales for a salesman each day is given below. What is the expected number of sales for this salesman? Round to the nearest tenth. n sales Probability of n sales Paragraph V 5 0.40 BI U A/ h 7 0.35 V + v 9 0.25 ...
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![Title: Probability Distribution and Expected Number of Sales
---
## Understanding Probability Distribution for Sales
The probability distribution for the number of sales a salesman makes each day is given in the table below. It allows us to calculate the expected number of sales for the salesman.
### Table: Probability Distribution for Number of Sales
| **n sales** | 5 | 7 | 9 |
|-----------------------|-----|-----|-----|
| **Probability of n sales** | 0.40 | 0.35 | 0.25 |
**Explanation:**
- The first row represents the number of sales the salesman can make in a day, denoted as \( n \).
- The second row lists the probability of achieving each corresponding number of sales.
### Calculating the Expected Number of Sales
The expected number of sales (E) can be calculated using the formula:
\[ E = \sum (n \times P(n)) \]
Where \( n \) is the number of sales and \( P(n) \) is the probability of \( n \) sales.
**Calculation:**
1. **For 5 sales:**
\[ 5 \times 0.40 = 2.00 \]
2. **For 7 sales:**
\[ 7 \times 0.35 = 2.45 \]
3. **For 9 sales:**
\[ 9 \times 0.25 = 2.25 \]
**Summation:**
\[ E = 2.00 + 2.45 + 2.25 = 6.70 \]
### Conclusion
The expected number of sales for the salesman each day is approximately **6.7**, rounded to the nearest tenth.
By understanding and utilizing probability distributions, we can make more informed decisions and predictions about future sales performance.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9d14dc9-6899-40d3-8d6b-066b6648b231%2F49f623fb-96f2-4e97-b281-fac82d9513d1%2F0ovoeq_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Probability Distribution and Expected Number of Sales
---
## Understanding Probability Distribution for Sales
The probability distribution for the number of sales a salesman makes each day is given in the table below. It allows us to calculate the expected number of sales for the salesman.
### Table: Probability Distribution for Number of Sales
| **n sales** | 5 | 7 | 9 |
|-----------------------|-----|-----|-----|
| **Probability of n sales** | 0.40 | 0.35 | 0.25 |
**Explanation:**
- The first row represents the number of sales the salesman can make in a day, denoted as \( n \).
- The second row lists the probability of achieving each corresponding number of sales.
### Calculating the Expected Number of Sales
The expected number of sales (E) can be calculated using the formula:
\[ E = \sum (n \times P(n)) \]
Where \( n \) is the number of sales and \( P(n) \) is the probability of \( n \) sales.
**Calculation:**
1. **For 5 sales:**
\[ 5 \times 0.40 = 2.00 \]
2. **For 7 sales:**
\[ 7 \times 0.35 = 2.45 \]
3. **For 9 sales:**
\[ 9 \times 0.25 = 2.25 \]
**Summation:**
\[ E = 2.00 + 2.45 + 2.25 = 6.70 \]
### Conclusion
The expected number of sales for the salesman each day is approximately **6.7**, rounded to the nearest tenth.
By understanding and utilizing probability distributions, we can make more informed decisions and predictions about future sales performance.
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