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
$$Mo = 30.5 + \left(\frac{14 - 12}{2(14) - 12 - 0}\right) \times 10$$
$$Mo = 30.5 + \left(\frac{2}{28 - 12}\right) \times 10$$
$$Mo = 30.5 + \left(\frac{2}{16}\right) \times 10$$
$$Mo = 30.5 + 1.25 = \mathbf{31.75}$$
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