SOLUTION: The weights (in kg) of 35 persons are given below:
43, 51, 47, 62, 48, 40, 50, 62, 53, 56, 40, 48, 56, 53, 50, 42, 55, 52, 48, 46, 45, 54, 52, 50, 47, 44, 54, 55, 60, 63, 58, 55,
Algebra.Com
Question 1166623: The weights (in kg) of 35 persons are given below:
43, 51, 47, 62, 48, 40, 50, 62, 53, 56, 40, 48, 56, 53, 50, 42, 55, 52, 48, 46, 45, 54, 52, 50, 47, 44, 54, 55, 60, 63, 58, 55, 60, 58, 53
a) Group this data by using the Sturges’ formula to determine the number of classes. Keep the width of each class equal.
b) Find the mean, median, standard deviation and variance of the grouped data obtained in (a) above.
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
This is a great exercise in statistical data grouping and analysis. Here are the steps to solve the problem.
## 📊 a) Grouping the Data
### 1. Find the Range ($R$)
The range is the difference between the highest and lowest values in the data set.
* **Highest Value ($H$):** 63 kg
* **Lowest Value ($L$):** 40 kg
* $$R = H - L = 63 - 40 = 23$$
### 2. Determine the Number of Classes ($k$) using Sturges' Formula
Sturges' formula is used to estimate the optimal number of class intervals.
$$k = 1 + 3.322 \log_{10}(n)$$
where $n$ is the total number of observations ($n = 35$).
$$k = 1 + 3.322 \log_{10}(35)$$
$$k \approx 1 + 3.322(1.544)$$
$$k \approx 1 + 5.132$$
$$k \approx 6.132$$
We round this number up to the nearest integer to get the number of classes.
$$\mathbf{k = 7 \text{ classes}}$$
### 3. Determine the Class Width ($w$)
The class width is the range divided by the number of classes.
$$w = \frac{R}{k} = \frac{23}{7} \approx 3.286$$
We must choose a convenient, round number slightly greater than 3.286 to ensure all data points are covered. Let's choose a class width of $\mathbf{4}$.
### 4. Create the Frequency Distribution Table
Starting the first class boundary at the lowest value, 40, and using a width of 4:
| Class Interval (Weight in kg) | Tally | Frequency ($f$) |
| :---: | :---: | :---: |
| 40 - 43 | IIII | 4 |
| 44 - 47 | IIII I | 6 |
| 48 - 51 | IIII IIII | 9 |
| 52 - 55 | IIII IIII | 9 |
| 56 - 59 | IIII | 5 |
| 60 - 63 | II | 2 |
| **Total** | | **35** |
---
## 💻 b) Find Mean, Median, Standard Deviation, and Variance
To calculate these statistics for **grouped data**, we must use the **class midpoint ($x$)** to represent the values in each interval.
| Class Interval | Midpoint ($x$) | Frequency ($f$) | $f \cdot x$ | $|x - \bar{x}|$ | $f \cdot |x - \bar{x}|$ | $x - \bar{x}$ | $(x - \bar{x})^2$ | $f \cdot (x - \bar{x})^2$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 40 - 43 | 41.5 | 4 | 166.0 | 9.07 | 36.28 | -9.07 | 82.26 | 329.04 |
| 44 - 47 | 45.5 | 6 | 273.0 | 5.07 | 30.42 | -5.07 | 25.70 | 154.20 |
| 48 - 51 | 49.5 | 9 | 445.5 | 1.07 | 9.63 | -1.07 | 1.14 | 10.26 |
| 52 - 55 | 53.5 | 9 | 481.5 | 2.93 | 26.37 | 2.93 | 8.58 | 77.22 |
| 56 - 59 | 57.5 | 5 | 287.5 | 6.93 | 34.65 | 6.93 | 47.98 | 239.90 |
| 60 - 63 | 61.5 | 2 | 123.0 | 10.93 | 21.86 | 10.93 | 119.46 | 238.92 |
| **Total** | | $\sum f = 35$ | $\sum fx = 1776.5$ | | | | | $\sum f(x - \bar{x})^2 = 1049.54$ |
### 1. Mean ($\bar{x}$)
The mean for grouped data is calculated as:
$$\bar{x} = \frac{\sum f x}{\sum f}$$
$$\bar{x} = \frac{1776.5}{35} \approx \mathbf{50.76 \text{ kg}}$$
### 2. Median ($M$)
The median is the value that falls at the $n/2$ position.
$$\frac{n}{2} = \frac{35}{2} = 17.5$$
The median class is the first class whose cumulative frequency is $\geq 17.5$.
* Cumulative Frequencies: 4, 10, **19**, 28, 33, 35.
* The median class is **48 - 51** (Cumulative Frequency 19).
Median Formula:
$$M = L + \left(\frac{\frac{n}{2} - C_f}{f_m}\right) \cdot w$$
Where:
* $L$: Lower boundary of the median class ($48 - 0.5 = 47.5$)
* $n$: Total frequency (35)
* $C_f$: Cumulative frequency of the class before the median class (10)
* $f_m$: Frequency of the median class (9)
* $w$: Class width (4)
$$M = 47.5 + \left(\frac{17.5 - 10}{9}\right) \cdot 4$$
$$M = 47.5 + \left(\frac{7.5}{9}\right) \cdot 4$$
$$M = 47.5 + 0.833 \cdot 4$$
$$M = 47.5 + 3.332 \approx \mathbf{50.83 \text{ kg}}$$
### 3. Variance ($\sigma^2$)
The variance is calculated using the formula:
$$\sigma^2 = \frac{\sum f(x - \bar{x})^2}{n - 1}$$
Using $n=35$:
$$\sigma^2 = \frac{1049.54}{35 - 1} = \frac{1049.54}{34} \approx \mathbf{30.87}$$
### 4. Standard Deviation ($\sigma$)
The standard deviation is the square root of the variance:
$$\sigma = \sqrt{\sigma^2}$$
$$\sigma = \sqrt{30.87} \approx \mathbf{5.56 \text{ kg}}$$
---
## 📝 Summary of Results
| Statistic | Value (Grouped Data) |
| :---: | :---: |
| **Mean** ($\bar{x}$) | $50.76$ kg |
| **Median** ($M$) | $50.83$ kg |
| **Variance** ($\sigma^2$) | $30.87$ |
| **Standard Deviation** ($\sigma$) | $5.56$ kg |
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