Statistics — Class 11 Maths Notes

Statistics · Class 11 Maths · 11 topics.

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Topics covered in Statistics

1.Introduction of Statistics
Statistics is a branch of mathematics that deals with collecting, analyzing, interpreting, presenting, and organizing data. It is widely used in various fields such as science, business, social sciences, and more. Let's dive into the basics of statistics with an example from everyday life.
What is Statistics?
Statistics helps us make sense of complex data by summarizing and interpreting it. There are two main branches of statistics:
1. Descriptive Statistics: This involves summarizing and describing the features of a dataset. It includes measures such as mean, median, mode, and standard deviation.
2. Inferential Statistics: This involves making predictions or inferences about a population based on a sample of data. It includes techniques like hypothesis testing, confidence intervals, and regression analysis.
Example of Descriptive Statistics
Imagine you have the scores of 10 students in a mathematics test 85, 78, 92, 88, 76, 95, 89, 81, 74, 90
Let's calculate some descriptive statistics for this dataset:
1. Mean (Average):
  Mean=Sum of all scoresNumber of scores=85+78+92+88+76+95+89+81+74+9010=84810=84.8Mean=Number of scoresSum of all scores=1085+78+92+88+76+95+89+81+74+90=10848=84.8
2. Median (Middle value when sorted): First, sort the scores:
  74,76,78,81,85,88,89,90,92,9574,76,78,81,85,88,89,90,92,95
  Since there are 10 scores (an even number), the median is the average of the 5th and 6th scores:
  Median=85+882=1732=86.5Median=285+88=2173=86.5
3. Mode (Most frequently occurring score): In this dataset, there is no score that repeats, so there is no mode.
4. Range (Difference between the highest and lowest scores):
  Range=95−74=21Range=95−74=21
5. Standard Deviation (A measure of the amount of variation in the scores): Standard deviation calculation involves several steps and is a bit more complex, but it gives an idea of how spread out the scores are around the mean.
Real-Life Application
Suppose you are a shop owner and you want to understand the average spending of your customers. By collecting data on how much each customer spends, you can calculate the mean, median, and mode to get a sense of typical spending patterns. This information can help you make decisions about inventory, pricing, and marketing strategies.
Careers Using Statistics
Statistics is crucial in many careers, such as:
- Data Analyst: Analyzing data to help companies make informed decisions.
- Market Researcher: Studying consumer behavior to guide product development and marketing.
- Actuary: Using statistical methods to assess risk in insurance and finance.
- Biostatistician: Applying statistics to biological and medical research.
Activity for Practice
1. Collect a dataset of your choice (e.g., daily temperatures for a month, scores in your favorite game, etc.).
2. Calculate the mean, median, mode, and range for your dataset.
3. Reflect on what these statistics tell you about your data.
2.Measures of Dispersion
Measures of dispersion describe the spread or variability within a set of data. They help us understand how much the data points differ from the mean or median. The most common measures of dispersion are range, variance, standard deviation, and interquartile range.
1. Range
The range is the simplest measure of dispersion. It is the difference between the highest and lowest values in a dataset.
Formula:
Range=Maximum Value−Minimum ValueRange=Maximum Value−Minimum Value
Example: Consider the following test scores: 85, 78, 92, 88, 76, 95, 89, 81, 74, 90.
- Maximum Value = 95
- Minimum Value = 74
Range=95−74=21Range=95−74=21
2. Variance
Variance measures the average squared deviation of each data point from the mean. It gives a sense of how spread out the data points are.
Formula for a sample:
𝑠2=∑(𝑥𝑖−𝑥‾)2𝑛−1s2=n−1∑(xi−x)2
where 𝑥𝑖xi is each data point, 𝑥‾x is the mean of the data, and 𝑛n is the number of data points.
Example: For the same test scores:
- Mean (𝑥‾x) = 84.8
- Calculate each squared deviation:(85−84.8)2,(78−84.8)2,(92−84.8)2,…(85−84.8)2,(78−84.8)2,(92−84.8)2,…
- Sum of squared deviations = 528.4
- Number of data points (n) = 10
𝑠2=528.410−1=528.49≈58.71s2=10−1528.4=9528.4≈58.71
3. Standard Deviation
Standard deviation is the square root of the variance. It provides a measure of dispersion in the same units as the data.
Formula:
𝑠=𝑠2s=s2
Example:
𝑠=58.71≈7.66s=58.71≈7.66
4. Interquartile Range (IQR)
The interquartile range measures the spread of the middle 50% of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1).
Formula:
IQR=𝑄3−𝑄1IQR=Q3−Q1
Example: For the test scores:
- Arrange the scores in ascending order: 74, 76, 78, 81, 85, 88, 89, 90, 92, 95.
- Q1 (first quartile) = 78
- Q3 (third quartile) = 90
IQR=90−78=12IQR=90−78=12
Real-Life Example of Dispersion
Suppose you are comparing the daily temperatures of two cities over a month. Even if both cities have the same average temperature, the variability (dispersion) might be different. One city might have a stable climate with little temperature change, while the other might experience significant fluctuations. Measures of dispersion will help you understand this variability.
Careers Using Measures of Dispersion
- Quality Control Analyst: Ensures product quality by monitoring variability in production processes.
- Economist: Analyzes economic data to understand market trends and variability.
- Meteorologist: Studies weather patterns and variability in climatic conditions.
- Psychometrician: Assesses the reliability and variability of psychological tests and measurements.
Activity for Practice
1. Collect a dataset of your choice (e.g., daily spending, number of steps walked each day, etc.).
2. Calculate the range, variance, standard deviation, and interquartile range for your dataset.
3. Interpret what these measures of dispersion tell you about your data.
3.Range: Understanding and Applications
Range is a simple yet important measure of dispersion that tells us the difference between the highest and lowest values in a dataset. It gives us a quick sense of the spread or variability within the data.
Formula:
Range=Maximum Value−Minimum ValueRange=Maximum Value−Minimum Value
Detailed Explanation
Let's break it down step-by-step with an everyday example.
Step-by-Step Explanation
1. Identify the Dataset: Suppose we have the test scores of a group of students:
  85,78,92,88,76,95,89,81,74,9085,78,92,88,76,95,89,81,74,90
2. Find the Maximum Value: The highest score in this dataset is 95.
3. Find the Minimum Value: The lowest score in this dataset is 74.
4. Calculate the Range:
  Range=95−74=21Range=95−74=21
The range of the test scores is 21, indicating the difference between the highest and lowest scores.
Real-Life Example
Imagine you are analyzing the daily temperatures of a city over a week. The temperatures (in Celsius) are:
30,32,35,33,31,29,3430,32,35,33,31,29,34
1. Maximum Temperature: 35°C
2. Minimum Temperature: 29°C
3. Range:Range=35−29=6Range=35−29=6
The range of the temperatures is 6°C, showing the temperature fluctuation during the week.
Problem-Solving Examples
Example 1: Monthly Rainfall
A meteorologist recorded the following rainfall (in mm) over a month:
120,85,95,110,90,130,115120,85,95,110,90,130,115
- Maximum Rainfall: 130 mm
- Minimum Rainfall: 85 mm
- Range:Range=130−85=45Range=130−85=45
The range of the monthly rainfall is 45 mm.
Example 2: Weekly Sales
A shopkeeper recorded the weekly sales (in units) for a product:
45,50,55,60,48,52,5845,50,55,60,48,52,58
- Maximum Sales: 60 units
- Minimum Sales: 45 units
- Range:Range=60−45=15Range=60−45=15
The range of the weekly sales is 15 units.
Example 3: Student Heights
The heights of students in a class (in cm) are:
150,160,155,170,165,158,162150,160,155,170,165,158,162
- Maximum Height: 170 cm
- Minimum Height: 150 cm
- Range:Range=170−150=20Range=170−150=20
The range of the student heights is 20 cm.
Practical Activity
1. Collect a dataset of your choice (e.g., the number of steps you walk each day for a week).
2. Identify the maximum and minimum values.
3. Calculate the range.
4. Reflect on what the range tells you about the variability in your dataset.
Careers Using Range
- Data Analyst: Uses range to quickly understand the spread of data.
- Meteorologist: Uses range to analyze temperature variations.
- Market Researcher: Uses range to study consumer spending habits.
4.Mean Deviation: Understanding and Applications
Mean Deviation (also known as Average Absolute Deviation) is a measure of dispersion that indicates the average of the absolute differences between each data point and the mean of the dataset. It provides insight into the variability or spread of the data around the mean.
Formula:
Mean Deviation=∑∣𝑥𝑖−𝑥‾∣𝑛Mean Deviation=n∑∣xi−x∣
where 𝑥𝑖xi is each data point, 𝑥‾x is the mean of the data, and 𝑛n is the number of data points.
Detailed Explanation
Let's break it down step-by-step with an everyday example.
Step-by-Step Explanation
1. Identify the Dataset: Suppose we have the test scores of a group of students:
  85,78,92,88,76,95,89,81,74,9085,78,92,88,76,95,89,81,74,90
2. Calculate the Mean:
  𝑥‾=∑𝑥𝑖𝑛=85+78+92+88+76+95+89+81+74+9010=84810=84.8x=n∑xi=1085+78+92+88+76+95+89+81+74+90=10848=84.8
3. Calculate the Absolute Deviations: Find the absolute difference between each data point and the mean:
  ∣85−84.8∣,∣78−84.8∣,∣92−84.8∣,∣88−84.8∣,∣76−84.8∣,∣95−84.8∣,∣89−84.8∣,∣81−84.8∣,∣74−84.8∣,∣90−84.8∣∣85−84.8∣,∣78−84.8∣,∣92−84.8∣,∣88−84.8∣,∣76−84.8∣,∣95−84.8∣,∣89−84.8∣,∣81−84.8∣,∣74−84.8∣,∣90−84.8∣0.2,6.8,7.2,3.2,8.8,10.2,4.2,3.8,10.8,5.20.2,6.8,7.2,3.2,8.8,10.2,4.2,3.8,10.8,5.2
4. Calculate the Mean Deviation:
  Mean Deviation=0.2+6.8+7.2+3.2+8.8+10.2+4.2+3.8+10.8+5.210=60.410=6.04Mean Deviation=100.2+6.8+7.2+3.2+8.8+10.2+4.2+3.8+10.8+5.2=1060.4=6.04
The mean deviation of the test scores is 6.04, indicating the average absolute deviation from the mean score.
Real-Life Example
Imagine you are analyzing the daily steps you walked over a week:
6000,7500,8200,5400,9100,6600,70006000,7500,8200,5400,9100,6600,7000
1. Calculate the Mean:
  𝑥‾=6000+7500+8200+5400+9100+6600+70007=498007≈7114.29x=76000+7500+8200+5400+9100+6600+7000=749800≈7114.29
2. Calculate the Absolute Deviations:
  ∣6000−7114.29∣,∣7500−7114.29∣,∣8200−7114.29∣,∣5400−7114.29∣,∣9100−7114.29∣,∣6600−7114.29∣,∣7000−7114.29∣∣6000−7114.29∣,∣7500−7114.29∣,∣8200−7114.29∣,∣5400−7114.29∣,∣9100−7114.29∣,∣6600−7114.29∣,∣7000−7114.29∣1114.29,385.71,1085.71,1714.29,1985.71,514.29,114.291114.29,385.71,1085.71,1714.29,1985.71,514.29,114.29
3. Calculate the Mean Deviation:
  Mean Deviation=1114.29+385.71+1085.71+1714.29+1985.71+514.29+114.297≈6894.297≈984.90Mean Deviation=71114.29+385.71+1085.71+1714.29+1985.71+514.29+114.29≈76894.29≈984.90
The mean deviation of the steps walked is approximately 984.90, indicating the average absolute deviation from the mean steps.
Problem-Solving Examples
Example 1: Monthly Rainfall
A meteorologist recorded the following rainfall (in mm) over a month:
120,85,95,110,90,130,115120,85,95,110,90,130,115
1. Calculate the Mean:
  𝑥‾=120+85+95+110+90+130+1157=7457≈106.43x=7120+85+95+110+90+130+115=7745≈106.43
2. Calculate the Absolute Deviations:
  ∣120−106.43∣,∣85−106.43∣,∣95−106.43∣,∣110−106.43∣,∣90−106.43∣,∣130−106.43∣,∣115−106.43∣∣120−106.43∣,∣85−106.43∣,∣95−106.43∣,∣110−106.43∣,∣90−106.43∣,∣130−106.43∣,∣115−106.43∣13.57,21.43,11.43,3.57,16.43,23.57,8.5713.57,21.43,11.43,3.57,16.43,23.57,8.57
3. Calculate the Mean Deviation:
  Mean Deviation=13.57+21.43+11.43+3.57+16.43+23.57+8.577≈98.577≈14.08Mean Deviation=713.57+21.43+11.43+3.57+16.43+23.57+8.57≈798.57≈14.08
Example 2: Weekly Sales
A shopkeeper recorded the weekly sales (in units) for a product:
45,50,55,60,48,52,5845,50,55,60,48,52,58
1. Calculate the Mean:
  𝑥‾=45+50+55+60+48+52+587=3687≈52.57x=745+50+55+60+48+52+58=7368≈52.57
2. Calculate the Absolute Deviations:
  ∣45−52.57∣,∣50−52.57∣,∣55−52.57∣,∣60−52.57∣,∣48−52.57∣,∣52−52.57∣,∣58−52.57∣∣45−52.57∣,∣50−52.57∣,∣55−52.57∣,∣60−52.57∣,∣48−52.57∣,∣52−52.57∣,∣58−52.57∣7.57,2.57,2.43,7.43,4.57,0.57,5.437.57,2.57,2.43,7.43,4.57,0.57,5.43
3. Calculate the Mean Deviation:
  Mean Deviation=7.57+2.57+2.43+7.43+4.57+0.57+5.437≈30.577≈4.37Mean Deviation=77.57+2.57+2.43+7.43+4.57+0.57+5.43≈730.57≈4.37
Example 3: Student Heights
The heights of students in a class (in cm) are:
150,160,155,170,165,158,162150,160,155,170,165,158,162
1. Calculate the Mean:
  𝑥‾=150+160+155+170+165+158+1627=11207≈160x=7150+160+155+170+165+158+162=71120≈160
2. Calculate the Absolute Deviations:
  ∣150−160∣,∣160−160∣,∣155−160∣,∣170−160∣,∣165−160∣,∣158−160∣,∣162−160∣∣150−160∣,∣160−160∣,∣155−160∣,∣170−160∣,∣165−160∣,∣158−160∣,∣162−160∣10,0,5,10,5,2,210,0,5,10,5,2,2
3. Calculate the Mean Deviation:
  Mean Deviation=10+0+5+10+5+2+27=347≈4.86Mean Deviation=710+0+5+10+5+2+2=734≈4.86
Practical Activity
1. Collect a dataset of your choice (e.g., the number of hours you study each day for a week).
2. Calculate the mean of your dataset.
3. Find the absolute deviations of each data point from the mean.
4. Calculate the mean deviation.
5. Reflect on what the mean deviation tells you about the variability in your dataset.
Careers Using Mean Deviation
- Quality Control Analyst: Uses mean deviation to monitor and maintain product quality.
- Economist: Analyzes economic data to understand variability and trends.
- Market Researcher: Uses mean deviation to study consumer behavior and spending patterns.
- Psychometrician: Assesses the reliability of psychological tests using mean deviation.
5.Mean Deviation for Ungrouped Data
Mean Deviation (also known as Average Absolute Deviation) for ungrouped data is a measure of dispersion that indicates the average of the absolute differences between each data point and the mean of the dataset. It provides insight into the variability or spread of the data around the mean.
Formula:
Mean Deviation=∑∣𝑥𝑖−𝑥‾∣𝑛Mean Deviation=n∑∣xi−x∣
where 𝑥𝑖xi is each data point, 𝑥‾x is the mean of the data, and 𝑛n is the number of data points.
Detailed Explanation
Let's break it down step-by-step with an everyday example.
Step-by-Step Explanation
1. Identify the Dataset: Suppose we have the daily number of hours a student studies over a week:
  2,3,4,5,6,3,42,3,4,5,6,3,4
2. Calculate the Mean:
  𝑥‾=∑𝑥𝑖𝑛=2+3+4+5+6+3+47=277≈3.86x=n∑xi=72+3+4+5+6+3+4=727≈3.86
3. Calculate the Absolute Deviations: Find the absolute difference between each data point and the mean:
  ∣2−3.86∣,∣3−3.86∣,∣4−3.86∣,∣5−3.86∣,∣6−3.86∣,∣3−3.86∣,∣4−3.86∣∣2−3.86∣,∣3−3.86∣,∣4−3.86∣,∣5−3.86∣,∣6−3.86∣,∣3−3.86∣,∣4−3.86∣1.86,0.86,0.14,1.14,2.14,0.86,0.141.86,0.86,0.14,1.14,2.14,0.86,0.14
4. Calculate the Mean Deviation:
  Mean Deviation=1.86+0.86+0.14+1.14+2.14+0.86+0.147≈7.147≈1.02Mean Deviation=71.86+0.86+0.14+1.14+2.14+0.86+0.14≈77.14≈1.02
The mean deviation of the study hours is approximately 1.02, indicating the average absolute deviation from the mean study hours.
Real-Life Example
Imagine you are analyzing the weekly grocery expenses (in INR) of a household over a month:
2000,2200,1800,25002000,2200,1800,2500
1. Calculate the Mean:
  𝑥‾=2000+2200+1800+25004=85004=2125x=42000+2200+1800+2500=48500=2125
2. Calculate the Absolute Deviations:
  ∣2000−2125∣,∣2200−2125∣,∣1800−2125∣,∣2500−2125∣∣2000−2125∣,∣2200−2125∣,∣1800−2125∣,∣2500−2125∣125,75,325,375125,75,325,375
3. Calculate the Mean Deviation:
  Mean Deviation=125+75+325+3754=9004=225Mean Deviation=4125+75+325+375=4900=225
The mean deviation of the weekly grocery expenses is 225, indicating the average absolute deviation from the mean expenses.
Problem-Solving Examples
Example 1: Monthly Electricity Consumption
A household records their monthly electricity consumption (in kWh) over six months:
300,320,290,310,330,300300,320,290,310,330,300
1. Calculate the Mean:
  𝑥‾=300+320+290+310+330+3006=18506≈308.33x=6300+320+290+310+330+300=61850≈308.33
2. Calculate the Absolute Deviations:
  ∣300−308.33∣,∣320−308.33∣,∣290−308.33∣,∣310−308.33∣,∣330−308.33∣,∣300−308.33∣∣300−308.33∣,∣320−308.33∣,∣290−308.33∣,∣310−308.33∣,∣330−308.33∣,∣300−308.33∣8.33,11.67,18.33,1.67,21.67,8.338.33,11.67,18.33,1.67,21.67,8.33
3. Calculate the Mean Deviation:
  Mean Deviation=8.33+11.67+18.33+1.67+21.67+8.336=696≈11.50Mean Deviation=68.33+11.67+18.33+1.67+21.67+8.33=669≈11.50
Example 2: Weekly Earnings
A freelancer records their weekly earnings (in INR) over five weeks:
4000,4500,4200,4800,46004000,4500,4200,4800,4600
1. Calculate the Mean:
  𝑥‾=4000+4500+4200+4800+46005=221005=4420x=54000+4500+4200+4800+4600=522100=4420
2. Calculate the Absolute Deviations:
  ∣4000−4420∣,∣4500−4420∣,∣4200−4420∣,∣4800−4420∣,∣4600−4420∣∣4000−4420∣,∣4500−4420∣,∣4200−4420∣,∣4800−4420∣,∣4600−4420∣420,80,220,380,180420,80,220,380,180
3. Calculate the Mean Deviation:
  Mean Deviation=420+80+220+380+1805=12805=256Mean Deviation=5420+80+220+380+180=51280=256
Example 3: Daily Water Consumption
A family records their daily water consumption (in liters) over a week:
100,120,110,130,125,115,105100,120,110,130,125,115,105
1. Calculate the Mean:
  𝑥‾=100+120+110+130+125+115+1057=8057≈115x=7100+120+110+130+125+115+105=7805≈115
2. Calculate the Absolute Deviations:
  ∣100−115∣,∣120−115∣,∣110−115∣,∣130−115∣,∣125−115∣,∣115−115∣,∣105−115∣∣100−115∣,∣120−115∣,∣110−115∣,∣130−115∣,∣125−115∣,∣115−115∣,∣105−115∣15,5,5,15,10,0,1015,5,5,15,10,0,10
3. Calculate the Mean Deviation:
  Mean Deviation=15+5+5+15+10+0+107=607≈8.57Mean Deviation=715+5+5+15+10+0+10=760≈8.57
Practical Activity
1. Collect a dataset of your choice (e.g., the number of hours you sleep each day for a week).
2. Calculate the mean of your dataset.
3. Find the absolute deviations of each data point from the mean.
4. Calculate the mean deviation.
5. Reflect on what the mean deviation tells you about the variability in your dataset.
Careers Using Mean Deviation
- Quality Control Analyst: Uses mean deviation to monitor and maintain product quality.
- Economist: Analyzes economic data to understand variability and trends.
- Market Researcher: Uses mean deviation to study consumer behavior and spending patterns.
- Psychometrician: Assesses the reliability of psychological tests using mean deviation.
6.Mean Deviation for Grouped Data
Mean Deviation can also be calculated for grouped data. There are two types of grouped data distributions: Discrete Frequency Distribution and Continuous Frequency Distribution. Let's go through the steps for calculating the mean deviation for each type.
(a) Mean Deviation for Discrete Frequency Distribution
In a discrete frequency distribution, data is presented in the form of distinct values and their corresponding frequencies.
Step-by-Step Explanation
1. Identify the Dataset: Let's consider a dataset of student scores with their frequencies:
  1. Calculate the Mean:
    𝑥‾=∑(𝑓⋅𝑥)∑𝑓=(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)2+3+5+4+1x=∑f∑(f⋅x)=2+3+5+4+1(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)𝑥‾=20+60+150+160+5015=44015≈29.33x=1520+60+150+160+50=15440≈29.33
  2. Calculate the Absolute Deviations: Find the absolute difference between each data point and the mean, then multiply by the frequency:
    1. Calculate the Mean Deviation:
      Mean Deviation=∑∣𝑥−𝑥‾∣⋅𝑓∑𝑓=38.66+28.99+3.35+42.68+20.6715=134.3515≈8.96Mean Deviation=∑f∑∣x−x∣⋅f=1538.66+28.99+3.35+42.68+20.67=15134.35≈8.96
    The mean deviation of the discrete frequency distribution is approximately 8.96.
    (b) Mean Deviation for Continuous Frequency Distribution
    In a continuous frequency distribution, data is presented in intervals with corresponding frequencies.
    Step-by-Step Explanation
    1. 5. Calculate the Mean Deviation:
      Mean Deviation=∑∣𝑚−𝑥‾∣⋅𝑓∑𝑓=102.5+189+21+256.5+156100=725100=7.25Mean Deviation=∑f∑∣m−x∣⋅f=100102.5+189+21+256.5+156=100725=7.25
    The mean deviation of the continuous frequency distribution is 7.25.
    Summary
    Mean Deviation for Discrete Frequency Distribution:
    1. Calculate the mean of the data.
    2. Find the absolute deviations from the mean.
    3. Multiply each deviation by the corresponding frequency.
    4. Sum the products and divide by the total frequency.
    Mean Deviation for Continuous Frequency Distribution:
    1. Calculate the class midpoints.
    2. Calculate the mean using midpoints and frequencies.
    3. Find the absolute deviations from the mean.
    4. Multiply each deviation by the corresponding frequency.
    5. Sum the products and divide by the total frequency.
7.Shortcut Method for Calculating Mean Deviation About Mean
The shortcut method simplifies the calculation of mean deviation, making it more efficient. This method is particularly useful when dealing with large datasets or grouped data.
(a) Mean Deviation for Discrete Frequency Distribution (Shortcut Method)
Step-by-Step Explanation
1. Identify the Dataset: Let's consider a dataset of student scores with their frequencies:
  Score (x)Frequency (f)102203305404501
2. Calculate the Mean (𝑥‾x):
  𝑥‾=∑(𝑓⋅𝑥)∑𝑓=(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)2+3+5+4+1x=∑f∑(f⋅x)=2+3+5+4+1(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)𝑥‾=20+60+150+160+5015=44015≈29.33x=1520+60+150+160+50=15440≈29.33
3. Calculate the Deviations and Absolute Deviations: Calculate the deviation (𝑑𝑖=𝑥𝑖−𝑥‾di=xi−x) for each value of 𝑥x and then the absolute deviation (∣𝑑𝑖∣∣di∣).
  Score (x)Frequency (f)𝑑𝑖=𝑥−𝑥‾∣𝑑𝑖∣10210−29.33=−19.3319.3320320−29.33=−9.339.3330530−29.33=0.670.6740440−29.33=10.6710.6750150−29.33=20.6720.67
4. Calculate ∣𝑑𝑖∣⋅𝑓∣di∣⋅f for each value:
  Score (x)Frequency (f)𝑑𝑖=𝑥−𝑥‾∣𝑑𝑖∣∣𝑑𝑖∣⋅𝑓102−19.3319.3319.33⋅2=38.66203−9.339.339.33⋅3=28.993050.670.670.67⋅5=3.3540410.6710.6710.67⋅4=42.6850120.6720.6720.67⋅1=20.67
5. Calculate the Mean Deviation:
  Mean Deviation=∑∣𝑑𝑖∣⋅𝑓∑𝑓=38.66+28.99+3.35+42.68+20.6715=134.3515≈8.96Mean Deviation=∑f∑∣di∣⋅f=1538.66+28.99+3.35+42.68+20.67=15134.35≈8.96
The mean deviation of the discrete frequency distribution is approximately 8.96.
(b) Mean Deviation for Continuous Frequency Distribution (Shortcut Method)
Step-by-Step Explanation
1. Identify the Dataset: Let's consider a dataset of student heights with their frequencies:
  Height (cm)Frequency (f)150−1595160−16918170−17942180−18927190−1998
2. Calculate the Class Midpoints (m):
  Midpoint=Lower limit+Upper limit2Midpoint=2Lower limit+Upper limitHeight (cm)Frequency (f)Midpoint (m)150−1595150+1592=154.5160−16918160+1692=164.5170−17942170+1792=174.5180−18927180+1892=184.5190−1998190+1992=194.5
3. Calculate the Mean (𝑥‾x):
  𝑥‾=∑(𝑓⋅𝑚)∑𝑓=(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)5+18+42+27+8x=∑f∑(f⋅m)=5+18+42+27+8(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)𝑥‾=772.5+2961+7329+4981.5+1556100=17500100=175x=100772.5+2961+7329+4981.5+1556=10017500=175
4. Calculate the Deviations and Absolute Deviations: Calculate the deviation (𝑑𝑖=𝑚𝑖−𝑥‾di=mi−x) for each midpoint and then the absolute deviation (∣𝑑𝑖∣∣di∣).
  Height (cm)Frequency (f)𝑑𝑖=𝑚−𝑥‾∣𝑑𝑖∣150−1595154.5−175=−20.520.5160−16918164.5−175=−10.510.5170−17942174.5−175=−0.50.5180−18927184.5−175=9.59.5190−1998194.5−175=19.519.5
5. Calculate ∣𝑑𝑖∣⋅𝑓∣di∣⋅f for each value:
  Height (cm)Frequency (f)𝑑𝑖=𝑚−𝑥‾∣𝑑𝑖∣∣𝑑𝑖∣⋅𝑓150−1595−20.520.520.5⋅5=102.5160−16918−10.510.510.5⋅18=189170−17942−0.50.50.5⋅42=21180−189279.59.59.5⋅27=256.5190−199819.519.519.5⋅8=156
6. Calculate the Mean Deviation:
  Mean Deviation=∑∣𝑑𝑖∣⋅𝑓∑𝑓=102.5+189+21+256.5+156100=725100=7.25Mean Deviation=∑f∑∣di∣⋅f=100102.5+189+21+256.5+156=100725=7.25
The mean deviation of the continuous frequency distribution is 7.25.
Summary
Shortcut Method for Discrete Frequency Distribution:
1. Calculate the mean (𝑥‾x).
2. Calculate the deviations (𝑑𝑖=𝑥𝑖−𝑥‾di=xi−x) and their absolute values (∣𝑑𝑖∣∣di∣).
3. Multiply each ∣𝑑𝑖∣∣di∣ by the corresponding frequency.
4. Sum these products and divide by the total frequency to get the mean deviation.
Shortcut Method for Continuous Frequency Distribution:
1. Calculate the class midpoints.
2. Calculate the mean (𝑥‾x) using midpoints and frequencies.
3. Calculate the deviations (𝑑𝑖=𝑚𝑖−𝑥‾di=mi−x) and their absolute values (∣𝑑𝑖∣∣di∣).
4. Multiply each ∣𝑑𝑖∣∣di∣ by the corresponding frequency.
5. Sum these products and divide by the total frequency to get the mean deviation
8.Mean Deviation About Median
Mean deviation about the median is a measure of dispersion that indicates the average of the absolute differences between each data point and the median of the dataset. It provides insight into the variability or spread of the data around the median.
Mean Deviation for Ungrouped Data
Step-by-Step Explanation
1. Identify the Dataset: Suppose we have a dataset of test scores:
  85,78,92,88,76,95,89,81,74,9085,78,92,88,76,95,89,81,74,90
2. Arrange the Data in Ascending Order:
  74,76,78,81,85,88,89,90,92,9574,76,78,81,85,88,89,90,92,95
3. Calculate the Median: Since the dataset has an even number of values (10 values), the median is the average of the 5th and 6th values:
  Median=85+882=1732=86.5Median=285+88=2173=86.5
4. Calculate the Absolute Deviations: Find the absolute difference between each data point and the median:
  ∣74−86.5∣,∣76−86.5∣,∣78−86.5∣,∣81−86.5∣,∣85−86.5∣,∣88−86.5∣,∣89−86.5∣,∣90−86.5∣,∣92−86.5∣,∣95−86.5∣∣74−86.5∣,∣76−86.5∣,∣78−86.5∣,∣81−86.5∣,∣85−86.5∣,∣88−86.5∣,∣89−86.5∣,∣90−86.5∣,∣92−86.5∣,∣95−86.5∣12.5,10.5,8.5,5.5,1.5,1.5,2.5,3.5,5.5,8.512.5,10.5,8.5,5.5,1.5,1.5,2.5,3.5,5.5,8.5
5. Calculate the Mean Deviation About the Median:
  Mean Deviation=∑∣𝑥𝑖−Median∣𝑛=12.5+10.5+8.5+5.5+1.5+1.5+2.5+3.5+5.5+8.510=6010=6Mean Deviation=n∑∣xi−Median∣=1012.5+10.5+8.5+5.5+1.5+1.5+2.5+3.5+5.5+8.5=1060=6
The mean deviation about the median of the test scores is 6.
Mean Deviation for Grouped Data
For grouped data, the process involves finding the median class and then using the class intervals and frequencies to calculate the mean deviation about the median.
(a) Mean Deviation for Discrete Frequency Distribution
Step-by-Step Explanation
1. Identify the Dataset: Let's consider a dataset of student scores with their frequencies:
  Score (x)Frequency (f)102203305404501
2. Calculate the Cumulative Frequency:
  Score (x)Frequency (f)Cumulative Frequency10222035305104041450115
3. Determine the Median: The median is the value corresponding to the cumulative frequency 𝑛22n. Here, 𝑛=15n=15, so 𝑛2=7.52n=7.5. The median class is 30 (since the cumulative frequency before 30 is 5 and after it is 10).
4. Calculate the Absolute Deviations:
  Score (x)Frequency (f)∣𝑥−Median∣⋅𝑓102∣10−30∣⋅2=40203∣20−30∣⋅3=30305∣30−30∣⋅5=0404∣40−30∣⋅4=40501∣50−30∣⋅1=20
5. Calculate the Mean Deviation About the Median:
  Mean Deviation=∑∣𝑥−Median∣⋅𝑓∑𝑓=40+30+0+40+2015=13015≈8.67Mean Deviation=∑f∑∣x−Median∣⋅f=1540+30+0+40+20=15130≈8.67
The mean deviation about the median of the discrete frequency distribution is approximately 8.67.
(b) Mean Deviation for Continuous Frequency Distribution
Step-by-Step Explanation
1. Identify the Dataset: Let's consider a dataset of student heights with their frequencies:
  Height (cm)Frequency (f)150−1595160−16918170−17942180−18927190−1998
2. Calculate the Cumulative Frequency:
  Height (cm)Frequency (f)Cumulative Frequency150−15955160−1691823170−1794265180−1892792190−1998100
3. Determine the Median Class: The median class is the one containing the 𝑛22nth value. Here, 𝑛=100n=100, so 𝑛2=502n=50. The median class is 170 - 179.
4. Calculate the Median:
  Median=𝐿+(𝑛2−𝐶𝐹𝑓)⋅ℎMedian=L+(f2n−CF)⋅h
  where 𝐿L is the lower boundary of the median class, 𝐶𝐹CF is the cumulative frequency of the class preceding the median class, 𝑓f is the frequency of the median class, and ℎh is the class width.
  Median=170+(50−2342)⋅10=170+(2742)⋅10≈170+6.43≈176.43Median=170+(4250−23)⋅10=170+(4227)⋅10≈170+6.43≈176.43
5. Calculate the Absolute Deviations: Calculate the deviation from the median for each class interval and multiply by the frequency:
  Height (cm)Frequency (f)Midpoint (m)∣𝑚−Median∣⋅𝑓150−1595154.5∣154.5−176.43∣⋅5=109.65160−16918164.5∣164.5−176.43∣⋅18=215.46170−17942174.5∣174.5−176.43∣⋅42=81.66180−18927184.5∣184.5−176.43∣⋅27=217.71190−1998194.5∣194.5−176.43∣⋅8=144.56
6. Calculate the Mean Deviation About the Median:
  Mean Deviation=∑∣𝑚−Median∣⋅𝑓∑𝑓=109.65+215.46+81.66+217.71+144.56100=769.04100≈7.69Mean Deviation=∑f∑∣m−Median∣⋅f=100109.65+215.46+81.66+217.71+144.56=100769.04≈7.69
The mean deviation about the median of the continuous frequency distribution is approximately 7.69.
Summary
Mean Deviation About the Median for Ungrouped Data:
1. Arrange the data in ascending order.
2. Calculate the median.
3. Find the absolute deviations from the median.
4. Calculate the mean of these absolute deviations.
Mean Deviation About the Median for Discrete Frequency Distribution:
1. Calculate the cumulative frequency.
2. Determine the median class.
3. Calculate the deviations from the median.
4. Find the absolute deviations and multiply by the frequency.
5. Calculate the mean deviation.
Mean Deviation About the Median for Continuous Frequency Distribution:
1. Calculate the cumulative frequency.
2. Determine the median class.
3. Calculate the median.
4. Calculate the deviations from the median for each class midpoint.
5. Find the absolute deviations and multiply by the frequency.
6. Calculate the mean deviation.
9.Limitations of Mean Deviation
Mean deviation is a useful measure of dispersion, but it has certain limitations that should be considered when using it to analyze data. Here are the key limitations:
1. Ignores Direction of Deviations:
  - Mean deviation only considers the absolute values of the deviations from the mean or median. It does not account for whether the deviations are positive or negative. This means it ignores the direction of the deviations, which can sometimes be important in data analysis.
2. Less Sensitive to Extreme Values:
  - Compared to other measures of dispersion like variance and standard deviation, mean deviation is less sensitive to extreme values (outliers). While this can be an advantage in some cases, it may also lead to underestimating the variability of the dataset if there are significant outliers.
3. Not Widely Used in Advanced Statistical Analysis:
  - Mean deviation is not as commonly used in advanced statistical analysis as variance and standard deviation. This is because variance and standard deviation have more desirable mathematical properties, such as being used in the calculation of other statistical measures (e.g., correlation, regression).
4. Does Not Minimize Squared Errors:
  - In statistical modeling and estimation, minimizing the sum of squared deviations is often more desirable. Variance and standard deviation are based on squared deviations, making them more appropriate for such purposes. Mean deviation, on the other hand, minimizes the sum of absolute deviations, which is less commonly used in optimization problems.
5. Lacks a Theoretical Basis in Probability Distributions:
  - Many probability distributions and statistical theories are based on the concept of variance and standard deviation rather than mean deviation. This makes mean deviation less applicable in theoretical contexts where these distributions are involved.
6. Not Suitable for All Types of Data:
  - Mean deviation is more suitable for data measured on an interval or ratio scale. It is less useful for ordinal or nominal data, where the concept of absolute deviation from a central value is less meaningful.
7. Complexity with Grouped Data:
  - Calculating mean deviation for grouped data can be more complex, especially if the data is not evenly distributed across intervals. This can make it less practical for large or highly variable datasets.
Summary of Limitations
- Ignores Direction of Deviations: Only considers absolute deviations, not the direction.
- Less Sensitive to Extreme Values: May underestimate variability in the presence of outliers.
- Not Widely Used in Advanced Analysis: Less commonly used than variance and standard deviation.
- Does Not Minimize Squared Errors: Minimizes absolute deviations, not squared deviations.
- Lacks Theoretical Basis: Less applicable in theoretical contexts involving probability distributions.
- Not Suitable for All Data Types: More suitable for interval or ratio data, less for ordinal or nominal data.
- Complexity with Grouped Data: More complex to calculate for grouped data, especially with uneven distribution.
10.Variance and Standard Deviation
Variance and standard deviation are fundamental statistical measures that quantify the spread or dispersion of a dataset. They are widely used in statistics to understand the variability within a dataset.
Variance
Variance measures the average squared deviations from the mean. It gives an idea of how much the values in a dataset deviate from the mean.
Formula for Variance (Ungrouped Data)
For a dataset 𝑥1,𝑥2,…,𝑥𝑛x1,x2,…,xn:
1. Calculate the Mean (𝑥‾x):
  𝑥‾=∑𝑖=1𝑛𝑥𝑖𝑛x=n∑i=1nxi
2. Calculate the Variance (𝑠2s2):
  𝑠2=∑𝑖=1𝑛(𝑥𝑖−𝑥‾)2𝑛s2=n∑i=1n(xi−x)2
Example
Consider the following dataset of test scores: 85,78,92,88,76,95,89,81,74,9085,78,92,88,76,95,89,81,74,90
1. Calculate the Mean (𝑥‾x):
  𝑥‾=85+78+92+88+76+95+89+81+74+9010=84810=84.8x=1085+78+92+88+76+95+89+81+74+90=10848=84.8
2. Calculate the Squared Deviations from the Mean:
  (85−84.8)2,(78−84.8)2,(92−84.8)2,(88−84.8)2,(76−84.8)2,(95−84.8)2,(89−84.8)2,(81−84.8)2,(74−84.8)2,(90−84.8)2(85−84.8)2,(78−84.8)2,(92−84.8)2,(88−84.8)2,(76−84.8)2,(95−84.8)2,(89−84.8)2,(81−84.8)2,(74−84.8)2,(90−84.8)20.04,46.24,52.84,10.24,76.84,104.04,17.64,14.44,118.24,27.040.04,46.24,52.84,10.24,76.84,104.04,17.64,14.44,118.24,27.04
3. Calculate the Variance:
  𝑠2=0.04+46.24+52.84+10.24+76.84+104.04+17.64+14.44+118.24+27.0410=467.610=46.76s2=100.04+46.24+52.84+10.24+76.84+104.04+17.64+14.44+118.24+27.04=10467.6=46.76
The variance of the test scores is 46.76.
Standard Deviation
Standard Deviation is the square root of the variance. It provides a measure of the average distance of each data point from the mean.
Formula for Standard Deviation (Ungrouped Data)
For a dataset 𝑥1,𝑥2,…,𝑥𝑛x1,x2,…,xn:
1. Calculate the Standard Deviation (𝑠s):𝑠=𝑠2=∑𝑖=1𝑛(𝑥𝑖−𝑥‾)2𝑛s=s2=n∑i=1n(xi−x)2
Example
Using the variance calculated above (𝑠2=46.76s2=46.76):
1. Calculate the Standard Deviation:𝑠=46.76≈6.84s=46.76≈6.84
The standard deviation of the test scores is approximately 6.84.
Variance and Standard Deviation for Grouped Data
For grouped data, we use the midpoints of the class intervals and their frequencies.
(a) Variance for Discrete Frequency Distribution
1. Identify the Dataset:
  Score (x)Frequency (f)102203305404501
2. Calculate the Mean (𝑥‾x):
  𝑥‾=∑(𝑓⋅𝑥)∑𝑓=(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)2+3+5+4+1=44015≈29.33x=∑f∑(f⋅x)=2+3+5+4+1(2⋅10)+(3⋅20)+(5⋅30)+(4⋅40)+(1⋅50)=15440≈29.33
3. Calculate the Squared Deviations from the Mean:
  Score (x)Frequency (f)(𝑥−𝑥‾)2(𝑥−𝑥‾)2⋅𝑓102(10−29.33)2=373.35373.35⋅2=746.7203(20−29.33)2=87.7887.78⋅3=263.34305(30−29.33)2=0.440.44⋅5=2.2404(40−29.33)2=113.44113.44⋅4=453.76501(50−29.33)2=427.78427.78⋅1=427.78
4. Calculate the Variance:
  𝑠2=∑(𝑓⋅(𝑥−𝑥‾)2)∑𝑓=746.7+263.34+2.2+453.76+427.7815=1893.7815≈126.25s2=∑f∑(f⋅(x−x)2)=15746.7+263.34+2.2+453.76+427.78=151893.78≈126.25
The variance of the discrete frequency distribution is approximately 126.25.
1. Calculate the Standard Deviation:𝑠=𝑠2=126.25≈11.24s=s2=126.25≈11.24
The standard deviation of the discrete frequency distribution is approximately 11.24.
(b) Variance for Continuous Frequency Distribution
1. Identify the Dataset:
  Height (cm)Frequency (f)150−1595160−16918170−17942180−18927190−1998
2. Calculate the Class Midpoints (m):
  Height (cm)Frequency (f)Midpoint (m)150−1595150+1592=154.5160−16918160+1692=164.5170−17942170+1792=174.5180−18927180+1892=184.5190−1998190+1992=194.5
3. Calculate the Mean (𝑥‾x):
  𝑥‾=∑(𝑓⋅𝑚)∑𝑓=(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)5+18+42+27+8x=∑f∑(f⋅m)=5+18+42+27+8(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)𝑥‾=772.5+2961+7329+4981.5+1556100=17500100=175x=100772.5+2961+7329+4981.5+1556=10017500=175
4. Calculate the Squared Deviations from the Mean:
  Height (cm)Frequency (f)(𝑚−𝑥‾)2(𝑚−𝑥‾)2⋅𝑓150−1595(154.5−175)2=420.25420.25⋅5=2101.25160−16918(164.5−175)2=110.25110.25⋅18=1984.5170−17942(174.5−175)2=0.250.25⋅42=10.5180−18927(184.5−175)2=90.2590.25⋅27=2436.75190−1998(194.5−175)2=380.25380.25⋅8=3042
5. Calculate the Variance:
  𝑠2=∑(𝑓⋅(𝑚−𝑥‾)2)∑𝑓=2101.25+1984.5+10.5+2436.75+3042100=9575100=95.75s2=∑f∑(f⋅(m−x)2)=1002101.25+1984.5+10.5+2436.75+3042=1009575=95.75
The variance of the continuous frequency distribution is 95.75.
1. Calculate the Standard Deviation:𝑠=𝑠2=95.75≈9.78s=s2=95.75≈9.78
The standard deviation of the continuous frequency distribution is approximately 9.78.
Summary
- Variance: Measures the average squared deviations from the mean.
- Standard Deviation: Square root of the variance; measures the average distance from the mean.
These measures are crucial for understanding the spread and variability in datasets, whether ungrouped or grouped.
11.Standard Deviation of a Continuous Frequency Distribution
The standard deviation for a continuous frequency distribution is calculated using the class intervals and their corresponding frequencies. Here’s a step-by-step method to calculate the standard deviation for a continuous frequency distribution.
Step-by-Step Explanation
1. Identify the Dataset: Consider a dataset of student heights (in cm) with their frequencies:
  Height (cm)Frequency (f)150−1595160−16918170−17942180−18927190−1998
2. Calculate the Class Midpoints (m): The midpoint (𝑚m) of each class interval is calculated as:
  𝑚=Lower limit+Upper limit2m=2Lower limit+Upper limitHeight (cm)Frequency (f)Midpoint (m)150−1595150+1592=154.5160−16918160+1692=164.5170−17942170+1792=174.5180−18927180+1892=184.5190−1998190+1992=194.5
3. Calculate the Mean (𝑥‾x): The mean (𝑥‾x) is calculated using the midpoints and frequencies:
  𝑥‾=∑(𝑓⋅𝑚)∑𝑓=(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)5+18+42+27+8x=∑f∑(f⋅m)=5+18+42+27+8(5⋅154.5)+(18⋅164.5)+(42⋅174.5)+(27⋅184.5)+(8⋅194.5)𝑥‾=772.5+2961+7329+4981.5+1556100=17500100=175x=100772.5+2961+7329+4981.5+1556=10017500=175
4. Calculate the Squared Deviations from the Mean: Calculate the squared deviation ((𝑚−𝑥‾)2(m−x)2) for each midpoint and multiply by the corresponding frequency:
  Height (cm)Frequency (f)(𝑚−𝑥‾)2(𝑚−𝑥‾)2⋅𝑓150−1595(154.5−175)2=420.25420.25⋅5=2101.25160−16918(164.5−175)2=110.25110.25⋅18=1984.5170−17942(174.5−175)2=0.250.25⋅42=10.5180−18927(184.5−175)2=90.2590.25⋅27=2436.75190−1998(194.5−175)2=380.25380.25⋅8=3042
5. Calculate the Variance (𝑠2s2): The variance (𝑠2s2) is calculated as:
  𝑠2=∑(𝑓⋅(𝑚−𝑥‾)2)∑𝑓=2101.25+1984.5+10.5+2436.75+3042100=9575100=95.75s2=∑f∑(f⋅(m−x)2)=1002101.25+1984.5+10.5+2436.75+3042=1009575=95.75
6. Calculate the Standard Deviation (𝑠s): The standard deviation (𝑠s) is the square root of the variance:
  𝑠=𝑠2=95.75≈9.78s=s2=95.75≈9.78
The standard deviation of the continuous frequency distribution is approximately 9.78.
Summary
- Class Midpoints: Calculate the midpoint of each class interval.
- Mean: Calculate the mean using the midpoints and frequencies.
- Squared Deviations: Calculate the squared deviations from the mean, multiplied by the frequencies.
- Variance: Sum these products and divide by the total frequency to find the variance.
- Standard Deviation: Take the square root of the variance to find the standard deviation.

Statistics — Class 11 Maths Notes

Topics covered in Statistics

1.Introduction of Statistics

What is Statistics?

Example of Descriptive Statistics

Real-Life Application

Careers Using Statistics

Activity for Practice

2.Measures of Dispersion

Measures of dispersion describe the spread or variability within a set of data. They help us understand how much the data points differ from the mean or median. The most common measures of dispersion are range, variance, standard deviation, and interquartile range.

1. Range

2. Variance

3. Standard Deviation

4. Interquartile Range (IQR)

Real-Life Example of Dispersion

Careers Using Measures of Dispersion

Activity for Practice

3.Range: Understanding and Applications

Range is a simple yet important measure of dispersion that tells us the difference between the highest and lowest values in a dataset. It gives us a quick sense of the spread or variability within the data.

Detailed Explanation

Step-by-Step Explanation

Real-Life Example

Problem-Solving Examples

Example 1: Monthly Rainfall

Example 2: Weekly Sales

Example 3: Student Heights

Practical Activity

Careers Using Range

4.Mean Deviation: Understanding and Applications

Mean Deviation (also known as Average Absolute Deviation) is a measure of dispersion that indicates the average of the absolute differences between each data point and the mean of the dataset. It provides insight into the variability or spread of the data around the mean.

Detailed Explanation

Step-by-Step Explanation

Real-Life Example

Problem-Solving Examples

Example 1: Monthly Rainfall

Example 2: Weekly Sales

Example 3: Student Heights

Practical Activity

Careers Using Mean Deviation

5.Mean Deviation for Ungrouped Data

Mean Deviation (also known as Average Absolute Deviation) for ungrouped data is a measure of dispersion that indicates the average of the absolute differences between each data point and the mean of the dataset. It provides insight into the variability or spread of the data around the mean.

Detailed Explanation

Step-by-Step Explanation

Real-Life Example

Problem-Solving Examples

Example 1: Monthly Electricity Consumption

Example 2: Weekly Earnings

Example 3: Daily Water Consumption

Practical Activity

Careers Using Mean Deviation

6.Mean Deviation for Grouped Data

(a) Mean Deviation for Discrete Frequency Distribution

Step-by-Step Explanation

(b) Mean Deviation for Continuous Frequency Distribution

Step-by-Step Explanation

Summary

7.Shortcut Method for Calculating Mean Deviation About Mean

The shortcut method simplifies the calculation of mean deviation, making it more efficient. This method is particularly useful when dealing with large datasets or grouped data.

(a) Mean Deviation for Discrete Frequency Distribution (Shortcut Method)

Step-by-Step Explanation

(b) Mean Deviation for Continuous Frequency Distribution (Shortcut Method)

Step-by-Step Explanation

Summary

8.Mean Deviation About Median

Mean deviation about the median is a measure of dispersion that indicates the average of the absolute differences between each data point and the median of the dataset. It provides insight into the variability or spread of the data around the median.

Mean Deviation for Ungrouped Data

Step-by-Step Explanation

Mean Deviation for Grouped Data

(a) Mean Deviation for Discrete Frequency Distribution

Step-by-Step Explanation

(b) Mean Deviation for Continuous Frequency Distribution

Step-by-Step Explanation

Summary

9.Limitations of Mean Deviation

Summary of Limitations

10.Variance and Standard Deviation

Variance and standard deviation are fundamental statistical measures that quantify the spread or dispersion of a dataset. They are widely used in statistics to understand the variability within a dataset.

Variance

Formula for Variance (Ungrouped Data)

Example