
# Grade 10 Compulsory Mathematics: Statistics Study Guide
## Unit: Statistics | Chapter: Mean Deviation | Complete Solutions

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## 🎯 Chapter Introduction: Understanding Mean Deviation

This guide focuses on **Mean Deviation**, another important tool for measuring the dispersion or spread of data. While quartile deviation only considers the middle 50% of data, mean deviation provides a more comprehensive picture by taking into account **every single data point**. It calculates the average distance of each data point from the center of the dataset (usually the mean or median). 📈

### Key Learning Areas
* **Defining Mean Deviation (MD):** Understanding what "average deviation" means.
* **Coefficient of Mean Deviation:** Learning how to compare the variability of different datasets.
* **Calculations from Mean:** Step-by-step methods to calculate MD using the arithmetic mean.
* **Calculations from Median:** Step-by-step methods to calculate MD using the median.
* **Choosing the Right Method:** Knowing when to use the mean versus the median.

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## 🔑 Key Concepts & Definitions

Let's break down the essential concepts before solving the problems.

### 1. What is Mean Deviation?
**Mean Deviation (MD)**, also called **Average Deviation**, is defined as the arithmetic average of the absolute differences (deviations) of each data point from a measure of central tendency. This central point can be the **mean**, **median**, or mode.

It essentially answers the question: "On average, how far is each data point from the center?" We use the **absolute value** (ignoring negative signs) because we only care about the distance, not the direction.

### 2. Coefficient of Mean Deviation
The **Coefficient of Mean Deviation** is a **relative measure** of dispersion. Because MD is an absolute measure (expressed in the same units as the data, like "marks" or "kg"), its coefficient is necessary for comparing the variability of two different datasets. It is a pure number without units.

* **Formula from Mean:**
    $$
    \text{Coefficient of MD from Mean} = \frac{\text{MD from Mean}}{\text{Mean}}
    $$
* **Formula from Median:**
    $$
    \text{Coefficient of MD from Median} = \frac{\text{MD from Median}}{\text{Median}}
    $$

### 3. Formulas & Calculation Methods

#### A. Mean Deviation from Mean
1.  **Calculate the Mean ($\bar{X}$):** Find the average of the dataset ($\bar{X} = \frac{\sum fx}{N}$).
2.  **Find Deviations:** Subtract the mean from each data point or mid-value ($D = x - \bar{X}$).
3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).
4.  **Calculate MD:** Find the average of these absolute deviations.
    $$
    \text{MD from Mean} = \frac{\sum f|D|}{N} = \frac{\sum f|x - \bar{X}|}{N}
    $$

#### B. Mean Deviation from Median
1.  **Calculate the Median ($M_d$):** Find the middle value of the dataset.
2.  **Find Deviations:** Subtract the median from each data point ($D = x - M_d$).
3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).
4.  **Calculate MD:** Find the average of these absolute deviations.
    $$
    \text{MD from Median} = \frac{\sum f|D|}{N} = \frac{\sum f|x - M_d|}{N}
    $$

---

## 📚 Exercise Solutions

### Short Questions

***

#### Question 1: Define mean deviation.
**Answer:**
Mean deviation is the average of the absolute values of the deviations of each item from a central point, which can be the mean, median, or mode.

***

#### Question 2: What is the coefficient of mean deviation?
**Answer:**
The coefficient of mean deviation is a relative measure of dispersion used to compare the variability of two or more series. It is the ratio of the mean deviation to the average (mean or median) from which it was calculated.

***

#### Question 3: Write the formulas for mean deviation and its coefficient from the mean.
**Answer:**
* **MD from Mean:** $MD = \frac{\sum f|m - \bar{X}|}{N}$
* **Coefficient of MD from Mean:** $\text{Coefficient} = \frac{\text{MD from Mean}}{\text{Mean}}$

***

#### Question 4: Write the formulas for mean deviation and its coefficient from the median.
**Answer:**
* **MD from Median:** $MD = \frac{\sum f|m - M_d|}{N}$
* **Coefficient of MD from Median:** $\text{Coefficient} = \frac{\text{MD from Median}}{\text{Median}}$

***

#### Question 5: Which method of calculating mean deviation is considered best and why?
**Answer:**
Calculating mean deviation from the **mean** is generally considered the best method. This is because the mean uses all the values in the dataset for its calculation, making it a more stable and representative measure of the center than the median, which is only a positional average.

***

#### Question 6: If $\sum fm = 1000$, $N = 50$, and $\sum f|m - \bar{X}| = 308$, find the MD from mean and its coefficient.
**Solution:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. Calculate Mean ($\bar{X}$)** | $\bar{X} = \frac{\sum fm}{N} = \frac{1000}{50} = \mathbf{20}$ |
| **2. Calculate MD from Mean** | $MD = \frac{\sum f|m - \bar{X}|}{N} = \frac{308}{50} = \mathbf{6.16}$ |
| **3. Calculate Coefficient of MD** | $\text{Coefficient} = \frac{MD}{\bar{X}} = \frac{6.16}{20} = \mathbf{0.308}$ |

***

#### Question 7: If $\sum f|m - \bar{X}| = 680$ and MD = 17, find $\sum f$.
**Solution:**

| Calculation Step | Formula & Result |
|:---|:---|
| **Find Total Frequency (N)** | $N = \frac{\sum f|m - \bar{X}|}{MD} = \frac{680}{17} = \mathbf{40}$ |

***

#### Question 8: If Median ($M_d$) = 40, $\sum f = 50$, and $\sum f|m - M_d| = 530$, find MD from median and its coefficient.
**Solution:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. Calculate MD from Median** | $MD = \frac{\sum f|m - M_d|}{N} = \frac{530}{50} = \mathbf{10.6}$ |
| **2. Calculate Coefficient of MD** | $\text{Coefficient} = \frac{MD}{M_d} = \frac{10.6}{40} = \mathbf{0.265}$ |

***

#### Question 9: If the coefficient of MD from median is 0.5 and the median is 40, find the MD.
**Solution:**

| Calculation Step | Formula & Result |
|:---|:---|
| **Find Mean Deviation (MD)** | $MD = \text{Coefficient} \times M_d = 0.5 \times 40 = \mathbf{20}$ |

---

### Long Questions (Worked Examples)

***

#### Question 10: Find the MD and its coefficient from the median for the data: 10, 50, 60, 40, 30, 20.
**Solution:**
1.  **Arrange data and find Median:**
    * Data: 10, 20, 30, 40, 50, 60.
    * $M_d = \frac{30 + 40}{2} = 35$.

2.  **Data Table:**

| X | D = &#124;X - 35&#124; |
|:---:|:---:|
| 10 | 25 |
| 20 | 15 |
| 30 | 5 |
| 40 | 5 |
| 50 | 15 |
| 60 | 25 |
| **Total** | **Σ&#124;D&#124; = 90** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. MD from Median** | $MD = \frac{\sum|D|}{N} = \frac{90}{6} = \mathbf{15}$ |
| **2. Coefficient of MD** | $\text{Coefficient} = \frac{MD}{M_d} = \frac{15}{35} \approx \mathbf{0.429}$ |

***

#### Question 11: Find the MD and its coefficient from the median.

| Marks (X) | Freq (f) |
|:---:|:---:|
| 20 | 2 |
| 30 | 4 |
| 40 | 5 |
| 50 | 7 |
| 60 | 2 |
| 70 | 1 |

**Solution:**
1.  **Find Median:** $N=21$. The median position is the $11^{th}$ item. From c.f. (2, 6, **11**,...), the $M_d = \mathbf{40}$.

2.  **Data Table:**

| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|
| 20 | 2 | 20 | 40 |
| 30 | 4 | 10 | 40 |
| 40 | 5 | 0 | 0 |
| 50 | 7 | 10 | 70 |
| 60 | 2 | 20 | 40 |
| 70 | 1 | 30 | 30 |
| **Total** | **N=21** | | **Σf&#124;D&#124;=220** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. MD from Median** | $MD = \frac{\sum f|D|}{N} = \frac{220}{21} \approx \mathbf{10.48}$ |
| **2. Coefficient of MD** | $\text{Coefficient} = \frac{MD}{M_d} = \frac{10.48}{40} \approx \mathbf{0.262}$ |

***

#### Question 12: Find the MD and its coefficient from the mean.

| Marks (X) | Freq (f) |
|:---:|:---:|
| 20 | 2 |
| 25 | 5 |
| 30 | 6 |
| 35 | 5 |
| 40 | 2 |

**Solution:**
1.  **Find Mean:** $N=20$. $\sum fX = 600$. $\bar{X} = \frac{600}{20} = \mathbf{30}$.

2.  **Data Table:**

| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|
| 20 | 2 | 10 | 20 |
| 25 | 5 | 5 | 25 |
| 30 | 6 | 0 | 0 |
| 35 | 5 | 5 | 25 |
| 40 | 2 | 10 | 20 |
| **Total** | **N=20** | | **Σf&#124;D&#124;=90** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. MD from Mean** | $MD = \frac{\sum f|D|}{N} = \frac{90}{20} = \mathbf{4.5}$ |
| **2. Coefficient of MD** | $\text{Coefficient} = \frac{MD}{\bar{X}} = \frac{4.5}{30} = \mathbf{0.15}$ |

***

#### Question 13: Calculate the MD from the median for the following data.

| Marks | Freq (f) |
|:---:|:---:|
| 10-20 | 5 |
| 20-30 | 4 |
| 30-40 | 5 |
| 40-50 | 4 |
| 50-60 | 2 |

**Solution:**
1.  **Find Median:** $N=20$. Median class is **30-40**.
    $M_d = L + \frac{\frac{N}{2} - c.f.}{f} \times i = 30 + \frac{10 - 9}{5} \times 10 = \mathbf{32}$.

2.  **Data Table:**

| Marks | m | f | D = &#124;m - 32&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|:---:|
| 10-20 | 15 | 5 | 17 | 85 |
| 20-30 | 25 | 4 | 7 | 28 |
| 30-40 | 35 | 5 | 3 | 15 |
| 40-50 | 45 | 4 | 13 | 52 |
| 50-60 | 55 | 2 | 23 | 46 |
| **Total** | | **N=20** | | **Σf&#124;D&#124;=226** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **MD from Median** | $MD = \frac{\sum f|D|}{N} = \frac{226}{20} = \mathbf{11.3}$ |

***

#### Question 14: Compute the mean deviation and its coefficient from the median.

| Marks (X) | Freq (f) |
|:---:|:---:|
| 55 | 4 |
| 65 | 5 |
| 75 | 6 |
| 85 | 3 |
| 95 | 2 |

**Solution:**
1.  **Find Median:** $N=20$. The median is the $10.5^{th}$ item. From c.f. (4, 9, **15**,...), the $M_d = \mathbf{75}$.

2.  **Data Table:**

| X | f | D = &#124;X - 75&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|
| 55 | 4 | 20 | 80 |
| 65 | 5 | 10 | 50 |
| 75 | 6 | 0 | 0 |
| 85 | 3 | 10 | 30 |
| 95 | 2 | 20 | 40 |
| **Total** | **N=20** | | **Σf&#124;D&#124;=200** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. MD from Median** | $MD = \frac{\sum f|D|}{N} = \frac{200}{20} = \mathbf{10}$ |
| **2. Coefficient of MD** | $\text{Coefficient} = \frac{MD}{M_d} = \frac{10}{75} \approx \mathbf{0.133}$ |

***

#### Question 15: Find the MD and its coefficient from the median.

| Marks (X) | Freq (f) |
|:---:|:---:|
| 10 | 2 |
| 20 | 3 |
| 30 | 9 |
| 40 | 21 |
| 50 | 11 |
| 60 | 5 |

**Solution:**
1.  **Find Median:** $N=51$. The median is the $26^{th}$ item. From c.f. (2, 5, 14, **35**,...), the $M_d = \mathbf{40}$.

2.  **Data Table:**

| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|
| 10 | 2 | 30 | 60 |
| 20 | 3 | 20 | 60 |
| 30 | 9 | 10 | 90 |
| 40 | 21 | 0 | 0 |
| 50 | 11 | 10 | 110 |
| 60 | 5 | 20 | 100 |
| **Total** | **N=51** | | **Σf&#124;D&#124;=420** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **1. MD from Median** | $MD = \frac{\sum f|D|}{N} = \frac{420}{51} \approx \mathbf{8.24}$ |
| **2. Coefficient of MD** | $\text{Coefficient} = \frac{MD}{M_d} = \frac{8.24}{40} \approx \mathbf{0.206}$ |

***

#### Question 16: Find the mean deviation from the median for the unsorted data below.

| Marks (X) | Freq (f) |
|:---:|:---:|
| 20 | 3 |
| 10 | 4 |
| 30 | 5 |
| 50 | 2 |
| 40 | 1 |

**Solution:**
1.  **Sort Data and Find Median:** $N=15$. The median is the $8^{th}$ item.

| X (sorted) | f | c.f. |
|:---:|:---:|:---:|
| 10 | 4 | 4 |
| 20 | 3 | 7 |
| 30 | 5 | **12** |
| 40 | 1 | 13 |
| 50 | 2 | 15 |

The 8th item falls in the class with mark **30**. So, $M_d = \mathbf{30}$.

2.  **Data Table:**

| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |
|:---:|:---:|:---:|:---:|
| 10 | 4 | 20 | 80 |
| 20 | 3 | 10 | 30 |
| 30 | 5 | 0 | 0 |
| 40 | 1 | 10 | 10 |
| 50 | 2 | 20 | 40 |
| **Total** | **N=15** | | **Σf&#124;D&#124;=160** |

3.  **Final Calculation:**

| Calculation Step | Formula & Result |
|:---|:---|
| **MD from Median** | $MD = \frac{\sum f|D|}{N} = \frac{160}{15} \approx \mathbf{10.67}$ |