Descriptive Statistics

Overview

In the study of engineering and computer science, we are frequently confronted with vast datasets, from system performance logs to experimental results. The ability to distill these collections of raw numbers into a meaningful and concise summary is a foundational pillar of quantitative analysis. Descriptive statistics provides the formal methodology for this process. It is the branch of statistics concerned with summarizing the features of a dataset, rather than using the data to infer properties about a larger population. The primary objective is to present quantitative descriptions in a manageable form, revealing underlying patterns and providing a clear, objective basis for interpretation.

This chapter introduces the fundamental tools required for this task. We shall explore two principal categories of descriptive measures: measures of central tendency, which identify the "center" or typical value of a distribution, and measures of dispersion, which quantify the "spread" or variability of the data points around that center. A firm command of these concepts is indispensable for the GATE examination, where questions frequently test one's ability to analyze given data distributions and compute their characteristic properties. Mastery of this material is not merely about calculation; it is about developing the analytical acumen to correctly interpret and compare datasets, a critical skill for any practicing engineer.

---

Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Measures of Central Tendency | Summarizing data with a single, central value. |
| 2 | Measures of Dispersion | Quantifying the spread or variability of data. |

---

Learning Objectives

---

We now turn our attention to Measures of Central Tendency...
## Part 1: Measures of Central Tendency

Introduction

In the study of statistics, our primary objective is often to distill a large set of observations into a few representative numerical values. When confronted with a dataset, whether it represents student marks, network packet delays, or processor clock cycles, we first seek a single value that can be considered the "center" or "typical" value of the distribution. Such a value provides a concise summary of the entire dataset, anchoring our understanding of its overall location on the number line.

The techniques for identifying this central point are known as measures of central tendency. These measures are fundamental to descriptive statistics, forming the bedrock upon which more complex analyses, such as measures of dispersion and inferential statistics, are built. For the GATE examination, a firm grasp of the three primary measures—the mean, the median, and the mode—is essential for interpreting data distributions and solving related quantitative problems. We shall explore the definition, calculation, and appropriate application of each of these measures for both ungrouped and grouped data.

---

Key Concepts

#
## 1. The Arithmetic Mean

The arithmetic mean, or simply the mean, is the most commonly used measure of central tendency. It is calculated by summing all the values in a dataset and dividing by the number of values. We can consider its calculation for two types of data presentations: ungrouped and grouped.

For Ungrouped Data:
If we have a set of $n$ observations $x_1, x_2, \dots, x_n$, the mean (denoted by $\bar{x}$) is their sum divided by the number of observations.

For Grouped Data:
When data is presented in a frequency distribution, we use the midpoint of each class interval as a representative value for that class.

Worked Example:

Problem: The marks obtained by 10 students in a test are: 15, 20, 25, 20, 15, 30, 45, 50, 20, 10. Calculate the mean mark.

Solution:

Step 1: Identify the observations and the total count.
The observations are $x_i = \{15, 20, 25, 20, 15, 30, 45, 50, 20, 10\}$.
The number of observations is $n = 10$.

Step 2: Sum all the observations.

Step 3: Compute the sum.

Step 4: Apply the formula for the mean.

Result:

Answer: The mean mark is $25$.

---

#
## 2. The Median

The median is the middle value of a dataset that has been arranged in ascending or descending order. It effectively divides the dataset into two equal halves. Unlike the mean, the median is not affected by extremely large or small values (outliers), making it a more robust measure of central tendency for skewed distributions.

For Ungrouped Data:
First, we must arrange the data in order.

• If the number of observations $n$ is odd, the median is the value at the $\left(\frac{n+1}{2}\right)^{th}$ position.


• If the number of observations $n$ is even, the median is the average of the two middle values, which are at the $\left(\frac{n}{2}\right)^{th}$ and $\left(\frac{n}{2} + 1\right)^{th}$ positions.

For Grouped Data:
For data in a frequency distribution, the median is found within the median class, which is the class interval where the cumulative frequency crosses $N/2$.

Worked Example:

Problem: Find the median of the following dataset: 9, 3, 5, 1, 8, 10, 6.

Solution:

Step 1: Arrange the data in ascending order.

Step 2: Determine the number of observations, $n$.
Here, $n = 7$, which is an odd number.

Step 3: Find the position of the median.
The position is $\left(\frac{n+1}{2}\right)^{th} = \left(\frac{7+1}{2}\right)^{th} = 4^{th}$ position.

Step 4: Identify the value at this position.
The value at the 4th position in the sorted list is 6.

Result:

Answer: The median of the dataset is $6$.

---

#
## 3. The Mode

The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), two modes (bimodal), more than two modes (multimodal), or no mode at all if all values appear with the same frequency.

For Grouped Data:
The mode lies in the modal class, which is the class interval with the highest frequency.

#
## 4. Relationship between Mean, Median, and Mode

For a perfectly symmetrical distribution, the mean, median, and mode are identical. However, for skewed distributions, these measures diverge. An empirical relationship exists for moderately skewed distributions.

The relationship can be visualized as follows:

Mode

Median

Mean
Positive Skew
(Mean > Median > Mode)

Mode

Median

Mean
Negative Skew
(Mean < Median < Mode)

---

Problem-Solving Strategies

---

Common Mistakes

---

Practice Questions

:::question type="NAT" question="The scores of a batsman in 10 innings are: 38, 70, 48, 34, 42, 55, 63, 46, 54, 44. The mean score is:" answer="49.4" hint="Sum all the scores and divide by the number of innings." solution="
Step 1: Sum the scores.

Step 2: Compute the sum.

Step 3: Divide by the number of innings, $n=10$.

Result:

"
:::

:::question type="MCQ" question="For a moderately skewed distribution, the mean is 30 and the mode is 24. What is the approximate value of the median?" options=["26", "27", "28", "29"] answer="28" hint="Use the empirical relationship: Mode ≈ 3 Median - 2 Mean." solution="
Step 1: State the empirical formula.

Step 2: Substitute the given values.

Step 3: Simplify the equation.

Step 4: Solve for the Median.

Result:

"
:::

:::question type="MCQ" question="What is the median of the dataset: 12, 4, 15, 7, 20, 9, 11?" options=["7", "9", "11", "12"] answer="11" hint="First, arrange the data in ascending order and then find the middle element." solution="
Step 1: Arrange the data in ascending order.

Step 2: Count the number of observations, $n$.
Here, $n=7$, which is an odd number.

Step 3: Find the position of the median term.
The position is $\left( \frac{n+1}{2} \right)^{th} = \left( \frac{7+1}{2} \right)^{th} = 4^{th}$.

Step 4: Identify the value at the 4th position.
The 4th value in the sorted list is 11.

Result:

"
:::

:::question type="MSQ" question="Which of the following statements about measures of central tendency are correct?" options=["The mean is sensitive to outliers.", "The median is always one of the data points in the dataset.", "The mode can be used for categorical data.", "For a positively skewed distribution, Mean > Median > Mode."] answer="The mean is sensitive to outliers.,The mode can be used for categorical data.,For a positively skewed distribution, Mean > Median > Mode." hint="Evaluate each statement's validity. Consider edge cases for the median." solution="

• The mean is sensitive to outliers: Correct. A single very large or very small value can significantly change the mean.


• The median is always one of the data points in the dataset: Incorrect. If there is an even number of data points, the median is the average of the two middle points, which may not be in the original dataset (e.g., median of {2, 4, 6, 8} is 5).


• The mode can be used for categorical data: Correct. The mode is the most frequent category (e.g., the most common color in a set of cars).


• For a positively skewed distribution, Mean > Median > Mode: Correct. The tail of the distribution is on the right, which pulls the mean to the right of the median.

"
:::

---

Summary

---

What's Next?

---

---

Part 2: Measures of Dispersion

Introduction

While measures of central tendency, such as the mean or median, provide a single value to represent the center of a dataset, they offer an incomplete picture of the data's characteristics. Two datasets may possess identical means yet exhibit vastly different distributions. To fully comprehend a dataset, we must also quantify the extent to which its values are spread out or clustered together. This quantification of spread is known as dispersion.

Measures of dispersion are statistical tools that describe the variability or scatter in a set of observations. A low measure of dispersion indicates that the data points tend to be clustered closely around the center, suggesting high uniformity. Conversely, a high measure of dispersion signifies that the data points are spread out over a wider range of values. In engineering and computer science, understanding dispersion is critical for applications ranging from performance analysis of algorithms to quality control in manufacturing processes.

---

Key Concepts

We will now examine the fundamental measures used to quantify dispersion.

#
## 1. Range

The range is the most straightforward measure of dispersion. It is defined as the difference between the maximum and minimum values in a dataset. While simple to compute, it is sensitive to outliers and considers only the two extreme values, ignoring the distribution of the remaining data points.

Worked Example:

Problem: Find the range for the dataset: $\{12, 15, 7, 22, 18, 5\}$.

Solution:

Step 1: Identify the maximum and minimum values.

Step 2: Apply the formula for the range.

Step 3: Compute the final answer.

Answer: The range of the dataset is $17$.

---

#
## 2. Variance and Standard Deviation

Variance and standard deviation are the most common and robust measures of dispersion. They quantify the average degree to which each point differs from the mean.

The variance is the average of the squared differences from the mean. Squaring the differences ensures that they are all positive and gives more weight to larger deviations. The standard deviation is simply the square root of the variance, which returns the measure of spread to the original units of the data, making it more interpretable.

μ
Large σ (High Dispersion)
Small σ (Low Dispersion)

Worked Example:

Problem: Calculate the population variance and standard deviation for the dataset: $\{2, 4, 4, 4, 5, 5, 7, 9\}$.

Solution:

Step 1: Calculate the population mean ($\mu$).

Step 2: Calculate the squared differences from the mean.

• $(2-5)^2 = (-3)^2 = 9$


• $(4-5)^2 = (-1)^2 = 1$


• $(4-5)^2 = (-1)^2 = 1$


• $(4-5)^2 = (-1)^2 = 1$


• $(5-5)^2 = (0)^2 = 0$


• $(5-5)^2 = (0)^2 = 0$


• $(7-5)^2 = (2)^2 = 4$


• $(9-5)^2 = (4)^2 = 16$

Step 3: Sum the squared differences.

Step 4: Calculate the variance ($\sigma^2$).

Step 5: Calculate the standard deviation ($\sigma$).

Answer: The population variance is $4$, and the standard deviation is $2$.

---

#
## 3. Coefficient of Variation (CV)

The standard deviation is an absolute measure of dispersion. It is expressed in the same units as the data. To compare the variability of two or more datasets with different means or different units, we require a relative measure of dispersion. The Coefficient of Variation is such a measure.

---

Problem-Solving Strategies

---

Common Mistakes

---

Practice Questions

:::question type="NAT" question="The scores of a student in 5 tests are 10, 12, 15, 13, 10. Calculate the population variance of these scores." answer="2.8" hint="First, find the mean of the scores. Then, use the formula for population variance: sum of squared differences from the mean, divided by the number of scores." solution="
Step 1: Calculate the mean ($\mu$) of the scores.

Step 2: Calculate the sum of the squared differences from the mean.

Step 3: Calculate the population variance ($\sigma^2$).

Wait, let me re-check my calculation.
Sum is 60. Mean is 12. Correct.
(10-12)^2 = 4
(12-12)^2 = 0
(15-12)^2 = 9
(13-12)^2 = 1
(10-12)^2 = 4
Sum of squares = 4+0+9+1+4 = 18. Correct.
Variance = 18/5 = 3.6.
Let me change the question or the answer. Let's change the numbers to get a cleaner answer.
Let the scores be 10, 8, 12, 10, 15.
Sum = 55. Mean = 11.
(10-11)^2 = 1
(8-11)^2 = 9
(12-11)^2 = 1
(10-11)^2 = 1
(15-11)^2 = 16
Sum of squares = 1+9+1+1+16 = 28.
Variance = 28/5 = 5.6.
Let's try another set: 6, 7, 8, 9, 10.
Mean = 8.
(6-8)^2 = 4
(7-8)^2 = 1
(8-8)^2 = 0
(9-8)^2 = 1
(10-8)^2 = 4
Sum of squares = 4+1+0+1+4 = 10.
Variance = 10/5 = 2.
This is a good question.
Let's re-create the original question with the answer 2.8
Let the scores be 4, 5, 6, 8, 7.
N=5. Sum = 30. Mean = 6.
(4-6)^2 = 4
(5-6)^2 = 1
(6-6)^2 = 0
(8-6)^2 = 4
(7-6)^2 = 1
Sum of squared diff = 4+1+0+4+1 = 10. Variance = 10/5 = 2.
Let's try to get 2.8.
Sum of squared diff / N = 2.8. If N=5, sum of squared diff = 14.
Let mean be $\mu$. We need $\sum(x_i-\mu)^2 = 14$.
Let's try scores: 10, 12, 15, 13, 10. Mean is 12. Sum of squared diff is 18. Variance is 3.6.
Let's stick with the numbers 6, 7, 8, 9, 10. The variance is 2.
Question: The scores of a student in 5 tests are 6, 7, 8, 9, 10. Calculate the population variance of these scores.
Answer: 2
Solution:
Step 1: Calculate the mean ($\mu$) of the scores.

Step 2: Calculate the sum of the squared differences from the mean.



Step 3: Calculate the population variance ($\sigma^2$).

Result: The population variance is 2.
This is a better question. Let's use this.
:::

:::question type="NAT" question="The scores of a student in 5 tests are 6, 7, 8, 9, 10. Calculate the population variance of these scores." answer="2" hint="First, find the mean of the scores. Then, use the formula for population variance: sum of squared differences from the mean, divided by the number of scores." solution="
Step 1: Calculate the mean ($\mu$) of the scores.

Step 2: Calculate the sum of the squared differences from the mean.

Step 3: Calculate the population variance ($\sigma^2$).

Result: The population variance is 2.
"
:::

:::question type="MCQ" question="A factory produces two types of microchips, A and B. For a batch of Chip A, the mean lifetime is 1000 hours with a standard deviation of 100 hours. For a batch of Chip B, the mean lifetime is 1200 hours with a standard deviation of 150 hours. Which of the following statements is correct regarding their relative variability?" options=["Chip A is more variable than Chip B","Chip B is more variable than Chip A","Both have the same variability","Variability cannot be compared"] answer="Chip B is more variable than Chip A" hint="To compare relative variability between datasets with different means, calculate the Coefficient of Variation (CV) for each." solution="
Step 1: Recall the formula for the Coefficient of Variation (CV).

Step 2: Calculate the CV for Chip A.

Step 3: Calculate the CV for Chip B.

Step 4: Compare the CV values.

Result: A higher CV indicates greater relative variability. Therefore, the lifetime of Chip B is more variable relative to its mean than that of Chip A.
"
:::

:::question type="MSQ" question="Which of the following statements about measures of dispersion are true?" options=["The standard deviation can be a negative value.","If all values in a dataset are identical, the variance is 0.","The range is highly sensitive to outliers.","The unit of variance is the same as the unit of the original data."] answer="If all values in a dataset are identical, the variance is 0.,The range is highly sensitive to outliers." hint="Analyze each statement based on the definitions of standard deviation, variance, and range." solution="

• Option A: The standard deviation is the square root of the variance. Variance is a sum of squared values, so it can never be negative. Thus, the standard deviation is always non-negative. This statement is false.


• Option B: If all values are identical, say $x_i = c$ for all $i$, then the mean $\mu = c$. Every term $(x_i - \mu)^2$ becomes $(c-c)^2 = 0$. The sum is 0, and therefore the variance is 0. This statement is true.


• Option C: The range is calculated as $x_{max} - x_{min}$. An extreme outlier would become either the new maximum or minimum, directly and significantly affecting the range. This statement is true.


• Option D: The variance is the average of squared deviations. Therefore, its unit is the square of the original data's unit (e.g., if data is in meters, variance is in meters-squared). The standard deviation has the same unit as the original data. This statement is false.

Result: The correct statements are that variance is 0 for an identical dataset and the range is sensitive to outliers.
"
:::
---

Summary

---

What's Next?

---

Chapter Summary

In this chapter, we have explored the fundamental principles of descriptive statistics, focusing on the methods used to summarize and describe the main features of a dataset. We began by examining measures of central tendency, which provide a single value to represent the center of a distribution. Following this, we investigated measures of dispersion, which quantify the extent to which the data points are spread out. A thorough understanding of these concepts is not merely foundational; it is essential for the quantitative analysis required in various engineering disciplines.

---

Chapter Review Questions

:::question type="MCQ" question="A dataset consists of the following five values: {10, 20, 30, 40, 150}. Which of the following statements is the most accurate description of the effect of the outlier (150) on the measures of central tendency?" options=["The mean will be significantly greater than the median.","The median will be significantly greater than the mean.","The mode will be the most representative measure of central tendency.","The mean and median will be approximately equal."] answer="A" hint="Consider how the arithmetic mean is calculated versus how the median is determined. Which one is more affected by extreme values?" solution="Let us analyze the dataset $S = \{10, 20, 30, 40, 150\}$.

Calculate the Median: The dataset is already sorted. The median is the middle value of the dataset. For a set of 5 items, the median is the 3rd item.

Calculate the Mean: The mean is the sum of the values divided by the count of the values.

Compare the Measures: We observe that the mean (50) is significantly greater than the median (30). This is a classic effect of a large positive outlier (a "right skew"). The outlier pulls the mean towards it, while the median, being position-based, is unaffected. The mode is not defined here as no value repeats. Therefore, the mean is inflated by the outlier and is much larger than the median.

"
:::

:::question type="NAT" question="The variance of a set of 20 data points is found to be 9. If each data point is first multiplied by 2 and then increased by 5, what is the new standard deviation of the resulting dataset?" answer="6" hint="Recall the effect of a linear transformation $Y = aX + b$ on the standard deviation. Does the additive constant 'b' affect the spread of the data?" solution="Let the original dataset be represented by the random variable $X$, and the new dataset by $Y$.

Given Information:

- The original variance is $\text{Var}(X) = \sigma_x^2 = 9$.
- The original standard deviation is the square root of the variance: $\sigma_x = \sqrt{9} = 3$.

Linear Transformation:

- Each data point is transformed according to the rule $Y = 2X + 5$.
- This is a linear transformation of the form $Y = aX + b$, with $a = 2$ and $b = 5$.

Effect on Standard Deviation:

- The properties of linear transformations on standard deviation state that $\sigma_y = |a|\sigma_x$.
- The additive constant $b$ shifts the entire dataset but does not change its spread, so it has no effect on the standard deviation or variance.

Calculation:

- Substitute the values of $a$ and $\sigma_x$:

- The new standard deviation is 6.
"
:::

:::question type="MCQ" question="For the dataset {8, 15, 12, 5, 22, 18, 10, 25}, what is the value of the Interquartile Range (IQR)?" options=["8","10","11.5","20"] answer="B" hint="First, sort the data. Then, find the median (Q2), followed by the median of the lower half (Q1) and the median of the upper half (Q3)." solution="

Sort the Dataset: First, we must arrange the data in ascending order.

The number of data points is $n=8$.

Find the Quartiles:

- First Quartile ($Q_1$): $Q_1$ is the median of the lower half of the data. The lower half is $\{5, 8, 10, 12\}$. Since there is an even number of points in this subset, $Q_1$ is the average of the two middle values.

- Third Quartile ($Q_3$): $Q_3$ is the median of the upper half of the data. The upper half is $\{15, 18, 22, 25\}$. Similarly, $Q_3$ is the average of the two middle values.

Calculate the IQR: The Interquartile Range is the difference between the third and first quartiles.

It appears there was a miscalculation in the thought process for the options. Let's re-evaluate based on a common alternative method for quartile calculation.

Alternative Method (Inclusive/Exclusive depends on convention, let's stick to the common textbook method above).
Let's re-check the calculation.
$S = \{5, 8, 10, 12, 15, 18, 22, 25\}$
Lower half: $\{5, 8, 10, 12\}$. Median $Q_1 = (8+10)/2 = 9$.
Upper half: $\{15, 18, 22, 25\}$. Median $Q_3 = (18+22)/2 = 20$.
IQR = $20 - 9 = 11$.

Textbook Author Note: There is a discrepancy between my calculated answer (11) and the provided options. This suggests an error in the question's options. Let me create a new question that works perfectly.

---
Revised Question for Integrity
:::question type="MCQ" question="For the dataset {2, 7, 5, 18, 15, 25, 10, 22}, what is the value of the Interquartile Range (IQR)?" options=["10", "15", "11", "17"] answer="C" hint="First, sort the data. Then, find the median (Q2), followed by the median of the lower half (Q1) and the median of the upper half (Q3)." solution="

Sort the Dataset: First, we must arrange the data in ascending order.

The number of data points is $n=8$.

Find the Quartiles:

- First Quartile ($Q_1$): $Q_1$ is the median of the lower half of the data. The lower half is $\{2, 5, 7, 10\}$. Since there is an even number of points in this subset, $Q_1$ is the average of the two middle values.

- Third Quartile ($Q_3$): $Q_3$ is the median of the upper half of the data. The upper half is $\{15, 18, 22, 25\}$. Similarly, $Q_3$ is the average of the two middle values.

Calculate the IQR: The Interquartile Range is the difference between the third and first quartiles.

Author Note 2: Another calculation error. Let's try again, very carefully. This is why proofreading is critical.
$Q_1 = (5+7)/2 = 6$.
$Q_3 = (18+22)/2 = 20$.
IQR = $20-6 = 14$. Let me create a question where the numbers work out cleanly with the options.

---
Final, Corrected Question
:::question type="MCQ" question="For the dataset {3, 9, 11, 15, 18, 21, 25, 30}, what is the value of the Interquartile Range (IQR)?" options=["10", "12", "13", "14.5"] answer="D" hint="First, sort the data. Then, find the median of the lower half (Q1) and the median of the upper half (Q3)." solution="

Sort the Dataset: The data is already sorted.

The number of data points is $n=8$.

Find the Quartiles:

- First Quartile ($Q_1$): $Q_1$ is the median of the lower half of the data: $\{3, 9, 11, 15\}$.

- Third Quartile ($Q_3$): $Q_3$ is the median of the upper half of the data: $\{18, 21, 25, 30\}$.

Calculate the IQR: The Interquartile Range is the difference between the third and first quartiles.

Author Note 3: My options are still wrong. I will fix the options to match the correct calculation.
The correct answer is 13. I will change option C to 13 and make that the answer. This is a realistic part of textbook creation.

---
Final, Actually Correct Question and Options
:::question type="MCQ" question="For the dataset {3, 9, 11, 15, 18, 21, 25, 30}, what is the value of the Interquartile Range (IQR)?" options=["10", "12", "13", "16.5"] answer="C" hint="First, sort the data. Then, find the median of the lower half (Q1) and the median of the upper half (Q3)." solution="

Sort the Dataset: The data is already sorted.

The number of data points is $n=8$.

Find the Quartiles:

- First Quartile ($Q_1$): $Q_1$ is the median of the lower half of the data: $\{3, 9, 11, 15\}$.

- Third Quartile ($Q_3$): $Q_3$ is the median of the upper half of the data: $\{18, 21, 25, 30\}$.

Calculate the IQR: The Interquartile Range is the difference between the third and first quartiles.

"
:::

---

What's Next?