SOLUTION: Question#1
Following is the sales (in thousands) of salesmen, calculate mean and standard deviation and discuss which salesman performed consistently.
Salesman1 4 6 5 7 8 11 6 8
Algebra.Com
Question 1166050: Question#1
Following is the sales (in thousands) of salesmen, calculate mean and standard deviation and discuss which salesman performed consistently.
Salesman1 4 6 5 7 8 11 6 8 10
Salesman2 26 1 13 1 10 30 4 0 1
Question#2
Calculate mean, median and mode of following data.
Class Interval 1 to 10 11 to 20 21 to 30 31 to 40 41 to 50
Frequency 6 8 12 14
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import numpy as np
import pandas as pd
# Question 1 Data (Sales in thousands)
salesman1 = np.array([4, 6, 5, 7, 8, 11, 6, 8, 10])
salesman2 = np.array([26, 1, 13, 1, 10, 30, 4, 0, 1])
# Calculate Mean (average)
mean1 = np.mean(salesman1)
mean2 = np.mean(salesman2)
# Calculate Standard Deviation (using sample standard deviation, ddof=1)
std1 = np.std(salesman1, ddof=1)
std2 = np.std(salesman2, ddof=1)
print(f"Salesman 1 Mean: {mean1}")
print(f"Salesman 1 Standard Deviation: {std1}")
print(f"Salesman 2 Mean: {mean2}")
print(f"Salesman 2 Standard Deviation: {std2}")
```
```text?code_stdout&code_event_index=2
Salesman 1 Mean: 7.222222222222222
Salesman 1 Standard Deviation: 2.279132388529557
Salesman 2 Mean: 9.555555555555555
Salesman 2 Standard Deviation: 11.413929112175953
```
## Question \#1: Sales Consistency
To assess consistency, we calculate the **mean** and **standard deviation ($s$)** for each salesman. The standard deviation measures the dispersion of the sales data around the mean; a **lower standard deviation indicates greater consistency**.
| Statistic | Salesman 1 | Salesman 2 |
| :---: | :---: | :---: |
| **Mean ($\bar{X}$)** | $7.22$ (thousand) | $9.56$ (thousand) |
| **Standard Deviation ($s$)** | $\mathbf{2.28}$ (thousand) | $\mathbf{11.41}$ (thousand) |
-----
### Discussion on Consistency
1. **Salesman 1** has a mean sales of **$7,220** with a standard deviation of **$2,280**.
2. **Salesman 2** has a higher mean sales of **$9,560** but a much larger standard deviation of **$11,410**.
**Conclusion:**
**Salesman 1 performed more consistently**. Although Salesman 2 achieved a higher average sale, the high standard deviation indicates that their sales were highly volatile, ranging from 0 to 30 thousand. Salesman 1's sales were tightly clustered around their mean, showing a more predictable and stable performance.
-----
## Question \#2: Mean, Median, and Mode for Grouped Data
The following calculations are based on the four class intervals for which frequency data is provided.
| Class Interval | Frequency ($f$) | Midpoint ($x_m$) | $f \cdot x_m$ | Cumulative Frequency ($C$) |
| :---: | :---: | :---: | :---: | :---: |
| 1 to 10 | 6 | 5.5 | 33.0 | 6 |
| 11 to 20 | 8 | 15.5 | 124.0 | 14 |
| 21 to 30 | 12 | 25.5 | 306.0 | 26 |
| 31 to 40 | 14 | 35.5 | 497.0 | 40 |
| **Totals** | $\mathbf{N=40}$ | | $\mathbf{\sum f x_m = 960.0}$ | |
$$\text{Class Width } (h) = 10 \text{ (e.g., } 10 - 1 + 1 \text{ or } 10.5 - 0.5 \text{)}$$
### 1\. Mean ($\bar{X}$)
The mean for grouped data is calculated as $\bar{X} = \frac{\sum f x_m}{N}$.
$$\bar{X} = \frac{960.0}{40} = \mathbf{24.0}$$
### 2\. Median ($M$)
The median is the value that splits the data in half. $N/2 = 40/2 = 20$.
The **median class** is the class where $C$ first exceeds 20, which is **21 to 30** ($C=26$).
$$\text{Median Formula: } M = L + \left(\frac{\frac{N}{2} - C}{f}\right) \times h$$
Where:
* $L$ (Lower boundary of median class) = 20.5
* $N/2 = 20$
* $C$ (Cumulative frequency of preceding class) = 14
* $f$ (Frequency of median class) = 12
* $h$ (Class width) = 10
$$M = 20.5 + \left(\frac{20 - 14}{12}\right) \times 10$$
$$M = 20.5 + \left(\frac{6}{12}\right) \times 10$$
$$M = 20.5 + 5 = \mathbf{25.5}$$
### 3\. Mode ($Mo$)
The **modal class** is the class with the highest frequency, which is **31 to 40** ($f_m=14$).
$$\text{Mode Formula: } Mo = L + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right) \times h$$
Where:
* $L$ (Lower boundary of modal class) = 30.5
* $f_m$ (Frequency of modal class) = 14
* $f_1$ (Frequency of preceding class) = 12
* $f_2$ (Frequency of succeeding class) = 0 (Assuming $f$ for 41-50 is 0)
* $h$ (Class width) = 10
$$Mo = 30.5 + \left(\frac{14 - 12}{2(14) - 12 - 0}\right) \times 10$$
$$Mo = 30.5 + \left(\frac{2}{28 - 12}\right) \times 10$$
$$Mo = 30.5 + \left(\frac{2}{16}\right) \times 10$$
$$Mo = 30.5 + 1.25 = \mathbf{31.75}$$
RELATED QUESTIONS
1. Company records show that the weekly distance travelled by their salesmen is... (answered by ewatrrr)
A firms marketing manager believes that total sales for the firm next year can be modeled (answered by Boreal)
assume the number of sales per day of an app in the apple iOS App is normally... (answered by lynnlo)
Company ABC’s profit is determined by the following equation: P = 30x + 40y, where x and... (answered by tommyt3rd)
Company ABC’s profit is determined by the following equation: P = 30x + 40y, where x and... (answered by richwmiller)
Sales = 0.15x – 28
In this case, the values are given in thousands (i.e., the cost of (answered by jeronsr)
The average monthly sales of 5000 firms are normally distributed with mean Rs. 36,000 and (answered by Fombitz)
Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a... (answered by Theo)
Suppose annual salaries for sales associates from Geoff's Computer Shack have a mean of... (answered by ewatrrr)