Class 10th Statistics Notes

14

Statistics

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Chapter Roadmap

1. Basic Terms of Statistics 2. Types and Organisation of Data 3. Mean of Grouped Data 4. Assumed Mean Method 5. Step-Deviation Method 6. Median of Grouped Data 7. Mode of Grouped Data 8. Empirical Relation 9. Board-Exam Question Patterns 10. 10 Detailed Solved Questions 11. Board Exam Strategy 12. Quick Revision

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Chapter Goal

This chapter teaches us how to organise numerical data and find its mean, median and mode. In board questions, the main skills are making a correct frequency table, finding class marks, locating the median class, identifying the modal class and applying the correct formula without calculation mistakes.

1. Basic Terms of Statistics

1.1 What is Statistics?

Statistics is the study of collecting, organising, presenting and analysing numerical data so that useful conclusions can be obtained.

1.2 Data

A collection of observations or numerical facts is called data. For example, the marks obtained by students in a test form a set of data.

Important Terms

Observation: Each individual value in a data set.

Frequency: The number of times an observation occurs.

Class interval: A range used to group observations, such as \(10-20\).

Class size: The difference between the upper and lower class boundaries for equal intervals.

Solved Example 1 — Find Frequency

Question: In the data \(2,3,3,4,3,5,2\), find the frequency of \(3\).

1. Count how many times \(3\) occurs.
2. The number \(3\) occurs three times.
3. Therefore its frequency is \(3\).
4. Answer: \(\boxed{3}\).

Solved Example 2 — Identify an Observation

Question: In the data \(5,8,11,14\), identify the observation \(11\).

1. The individual values in a data set are observations.
2. \(11\) is one of the given values.
3. Therefore \(11\) is an observation.
4. Answer: \(\boxed{11}\).

Solved Example 3 — Class Interval

Question: What is the class interval in \(20-30\)?

1. The lower limit is \(20\).
2. The upper limit is \(30\).
3. So the class interval is \(20-30\).
4. Its size is \(30-20=10\).
5. Answer: Class size \(=\boxed{10}\).

2. Types and Organisation of Data

2.1 Raw Data

Data given in its original, unorganised form is called raw data.

2.2 Frequency Distribution

A frequency distribution arranges observations into suitable groups and records how many observations fall in each group.

Class Interval	Frequency
0–10	4
10–20	7
20–30	9

2.3 Class Mark

The class mark is the midpoint of a class interval.

\[ \boxed{x_i=\frac{\text{Lower limit}+\text{Upper limit}}{2}} \]

Solved Example 1 — Class Mark

Question: Find the class mark of \(10-20\).

1. Use \(x_i=\frac{\text{lower limit}+\text{upper limit}}{2}\).
2. \(x_i=\frac{10+20}{2}\).
3. \(x_i=\frac{30}{2}=15\).
4. Answer: \(\boxed{15}\).

Solved Example 2 — Class Mark

Question: Find the class mark of \(30-40\).

1. Add the limits: \(30+40=70\).
2. Divide by \(2\): \(70/2=35\).
3. Answer: \(\boxed{35}\).

Solved Example 3 — Class Size

Question: Find the class size of \(40-50\).

1. Class size \(=50-40\).
2. \(=10\).
3. Answer: \(\boxed{10}\).

3. Mean of Grouped Data – Direct Method

For grouped data, first find the class mark \(x_i\) of every class.

\[ \boxed{\bar{x}=\frac{\sum f_i x_i}{\sum f_i}} \]

Here \(f_i\) is the frequency, \(x_i\) is the class mark and \(\sum f_i\) is the total frequency.

3.1 Step-by-Step Method

1. Find the class mark \(x_i\) for every class.
2. Multiply each \(x_i\) by its frequency \(f_i\).
3. Find \(\sum f_i x_i\).
4. Find \(\sum f_i\).
5. Use \(\bar{x}=\frac{\sum f_ix_i}{\sum f_i}\).

Solved Example 1 — Direct Method

Question: Find the mean of the following distribution.

Class	\(f_i\)
0–10	2
10–20	3
20–30	5

1. Class marks are \(5,15,25\).
2. Calculate \(f_ix_i\): \(2(5)=10,\ 3(15)=45,\ 5(25)=125\).
3. \(\sum f_i=2+3+5=10\).
4. \(\sum f_ix_i=10+45+125=180\).
5. \(\bar{x}=180/10=18\).
6. Answer: \(\boxed{18}\).

Solved Example 2 — Direct Method

Question: Find the mean for class marks \(10,20,30\) with frequencies \(2,4,4\).

1. Find \(f_ix_i\): \(20,80,120\).
2. \(\sum f_i=2+4+4=10\).
3. \(\sum f_ix_i=20+80+120=220\).
4. \(\bar{x}=220/10=22\).
5. Answer: \(\boxed{22}\).

Solved Example 3 — Mean from Table

Question: Class intervals are \(0-10,10-20,20-30\) and frequencies are \(4,3,3\). Find the mean.

1. Class marks: \(5,15,25\).
2. \(f_ix_i=20,45,75\).
3. \(\sum f_i=10\).
4. \(\sum f_ix_i=140\).
5. \(\bar{x}=140/10=14\).
6. Answer: \(\boxed{14}\).

4. Assumed Mean Method

When the class marks are large, the direct method can involve lengthy multiplication. The assumed mean method makes the arithmetic easier.

\[ \boxed{\bar{x}=a+\frac{\sum f_id_i}{\sum f_i}} \] \[ \boxed{d_i=x_i-a} \]

Here \(a\) is the assumed mean and \(d_i\) is the deviation from the assumed mean.

Solved Example 1 — Assumed Mean

Question: Find the mean when \(x_i=10,20,30\) and \(f_i=2,3,5\). Take \(a=20\).

1. Calculate deviations: \(d_i=x_i-a\).
2. So \(d_i=-10,0,10\).
3. Calculate \(f_id_i=-20,0,50\).
4. \(\sum f_i=10\), and \(\sum f_id_i=30\).
5. \(\bar{x}=20+\frac{30}{10}=23\).
6. Answer: \(\boxed{23}\).

Solved Example 2 — Choose an Assumed Mean

Question: For class marks \(15,25,35,45\), suggest a convenient assumed mean.

1. Choose a value near the centre.
2. \(35\) is one of the middle class marks.
3. Using \(35\) makes deviations simple: \(-20,-10,0,10\).
4. Answer: A convenient assumed mean is \(\boxed{35}\).

Solved Example 3 — Mean

Question: Find the mean for \(x_i=20,30,40\) and \(f_i=3,4,3\), taking \(a=30\).

1. Deviations: \(-10,0,10\).
2. \(f_id_i=-30,0,30\).
3. \(\sum f_id_i=0\).
4. \(\sum f_i=10\).
5. \(\bar{x}=30+\frac{0}{10}=30\).
6. Answer: \(\boxed{30}\).

5. Step-Deviation Method

The step-deviation method is especially useful when class sizes are equal and deviations have a common factor.

\[ \boxed{\bar{x}=a+h\frac{\sum f_iu_i}{\sum f_i}} \] \[ \boxed{u_i=\frac{x_i-a}{h}} \]

Here \(a\) is the assumed mean and \(h\) is the common class size.

Solved Example 1 — Step Deviation

Question: For \(x_i=10,20,30\), \(f_i=2,3,5\), take \(a=20\) and \(h=10\). Find the mean.

1. \(u_i=(x_i-a)/h\).
2. Thus \(u_i=-1,0,1\).
3. \(f_iu_i=-2,0,5\).
4. \(\sum f_iu_i=3\), \(\sum f_i=10\).
5. \(\bar{x}=20+10(3/10)=23\).
6. Answer: \(\boxed{23}\).

Solved Example 2 — Identify \(h\)

Question: What is \(h\) if class intervals are \(0-10,10-20,20-30\)?

1. Find the difference between consecutive limits.
2. \(10-0=10\).
3. Therefore the common class size is \(10\).
4. Answer: \(\boxed{h=10}\).

Solved Example 3 — Quick Mean

Question: Class marks are \(5,15,25\), frequencies \(1,2,2\). Take \(a=15,\ h=10\).

1. \(u_i=-1,0,1\).
2. \(f_iu_i=-1,0,2\).
3. \(\sum f_iu_i=1\), \(\sum f_i=5\).
4. \(\bar{x}=15+10(1/5)=17\).
5. Answer: \(\boxed{17}\).

6. Median of Grouped Data

The median is the middle value of a data set when the observations are arranged in order. For grouped data, we first find the cumulative frequency.

\[ \boxed{\text{Median}=l+\left(\frac{\frac n2-cf}{f}\right)h} \]

Here:

• \(l\) = lower boundary of the median class
• \(n\) = total frequency
• \(cf\) = cumulative frequency of the class just before the median class
• \(f\) = frequency of the median class
• \(h\) = class size

6.1 How to Find the Median Class

1. Find \(n=\sum f\).
2. Calculate \(n/2\).
3. Prepare cumulative frequency.
4. The class whose cumulative frequency is just greater than \(n/2\) is the median class.

Solved Example 1 — Find Median

Question: Find the median of the distribution below.

Class	\(f\)	cf
0–10	5	5
10–20	8	13
20–30	12	25
30–40	5	30

1. Total \(n=30\).
2. \(n/2=15\).
3. The first cf greater than \(15\) is \(25\), so median class is \(20-30\).
4. Here \(l=20,\ cf=13,\ f=12,\ h=10\).
5. \(\text{Median}=20+\left(\frac{15-13}{12}\right)10\).
6. \(=20+\frac{20}{12}=21.67\) approximately.
7. Answer: \(\boxed{21.67\text{ (approximately)}}\).

Solved Example 2 — Locate Median Class

Question: If \(n=50\) and cumulative frequencies are \(7,18,32,45,50\), find the median class position.

1. \(n/2=25\).
2. Look for the first cumulative frequency greater than \(25\).
3. It is \(32\).
4. Therefore the class corresponding to cf \(32\) is the median class.
5. Answer: The median class is the class with \(\boxed{cf=32}\).

Solved Example 3 — Median Formula

Question: In a grouped distribution, \(l=30,\ n=40,\ cf=17,\ f=8,\ h=10\). Find the median.

1. \(n/2=20\).
2. \(\text{Median}=30+\left(\frac{20-17}{8}\right)10\).
3. \(=30+\frac{30}{8}\).
4. \(=33.75\).
5. Answer: \(\boxed{33.75}\).

7. Mode of Grouped Data

The mode is the value that occurs most frequently. For grouped data, the class having the highest frequency is called the modal class.

\[ \boxed{\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h} \]

Here \(l\) is the lower boundary of the modal class, \(f_1\) is its frequency, \(f_0\) is the frequency of the class before it, \(f_2\) is the frequency of the class after it, and \(h\) is the class size.

Solved Example 1 — Find Modal Class

Question: Frequencies are \(4,7,12,8,5\). Which class is modal?

1. Find the highest frequency.
2. The highest frequency is \(12\).
3. The class corresponding to \(12\) is the modal class.
4. Answer: The class with frequency \(\boxed{12}\) is the modal class.

Solved Example 2 — Calculate Mode

Question: For a distribution, \(l=20,\ f_1=12,\ f_0=8,\ f_2=6,\ h=10\). Find the mode.

1. \(\text{Mode}=20+\left(\frac{12-8}{2(12)-8-6}\right)10\).
2. \(=20+\left(\frac4{10}\right)10\).
3. \(=20+4=24\).
4. Answer: \(\boxed{24}\).

Solved Example 3 — Identify \(f_0,f_1,f_2\)

Question: If the modal class is \(30-40\), what do \(f_0,f_1,f_2\) represent?

1. \(f_1\) is the frequency of the modal class \(30-40\).
2. \(f_0\) is the frequency of the preceding class \(20-30\).
3. \(f_2\) is the frequency of the succeeding class \(40-50\).
4. Answer: \(f_0\) = previous, \(f_1\) = modal, \(f_2\) = next frequency.

8. Empirical Relation Between Mean, Median and Mode

For a moderately skewed distribution, an important empirical relationship is:

\[ \boxed{\text{Mode}=3(\text{Median})-2(\text{Mean})} \]

It can also be rearranged as:

\[ \boxed{\text{Median}=\frac{\text{Mode}+2(\text{Mean})}{3}} \]

Solved Example 1 — Find Mode

Question: Mean \(=20\), Median \(=22\). Find the mode.

1. Use \(\text{Mode}=3\text{Median}-2\text{Mean}\).
2. \(=3(22)-2(20)\).
3. \(=66-40=26\).
4. Answer: \(\boxed{26}\).

Solved Example 2 — Find Median

Question: Mean \(=18\), Mode \(=24\). Find the median.

1. Use \(\text{Median}=\frac{\text{Mode}+2\text{Mean}}3\).
2. \(=\frac{24+2(18)}3\).
3. \(=\frac{60}{3}=20\).
4. Answer: \(\boxed{20}\).

Solved Example 3 — Check Relation

Question: Mean \(=15\), Median \(=16\), Mode \(=18\). Check the empirical relation.

1. Calculate \(3(\text{Median})-2(\text{Mean})\).
2. \(=3(16)-2(15)=48-30=18\).
3. This equals the given mode.
4. Answer: The relation is satisfied.

9. Important Board-Exam Question Patterns

Pattern 1 — Mean by Direct Method

Find class marks, calculate \(f_ix_i\), then use \(\bar{x}=\frac{\sum f_ix_i}{\sum f_i}\).

Pattern 2 — Mean by Assumed Mean / Step-Deviation

Choose a convenient \(a\), calculate deviations and substitute in the appropriate formula.

Pattern 3 — Median

Find total frequency, calculate \(n/2\), make cumulative frequency and identify the median class before applying the formula.

Pattern 4 — Mode

Identify the modal class from the highest frequency and carefully select \(f_0,f_1,f_2\).

10. 10 Detailed Solved Questions

Question 1 — Class Mark

Find the class marks of \(10-20,20-30,30-40\).

1. For \(10-20\): \(x=\frac{10+20}{2}=15\).
2. For \(20-30\): \(x=\frac{20+30}{2}=25\).
3. For \(30-40\): \(x=\frac{30+40}{2}=35\).
4. Answer: \(\boxed{15,25,35}\).

Question 2 — Mean Direct Method

Find the mean for \(x_i=10,20,30\) and \(f_i=3,2,5\).

1. \(f_ix_i=30,40,150\).
2. \(\sum f_i=10\).
3. \(\sum f_ix_i=220\).
4. \(\bar{x}=220/10=22\).
5. Answer: \(\boxed{22}\).

Question 3 — Mean Assumed Mean

Find the mean for \(x_i=15,25,35,45\) and \(f_i=2,3,4,1\), taking \(a=35\).

1. Deviations \(d_i=-20,-10,0,10\).
2. \(f_id_i=-40,-30,0,10\).
3. \(\sum f_id_i=-60\).
4. \(\sum f_i=10\).
5. \(\bar{x}=35+\frac{-60}{10}=29\).
6. Answer: \(\boxed{29}\).

Question 4 — Step-Deviation

For \(x_i=10,20,30,40\) and \(f_i=2,4,3,1\), take \(a=20,\ h=10\). Find the mean.

1. \(u_i=-1,0,1,2\).
2. \(f_iu_i=-2,0,3,2\).
3. \(\sum f_iu_i=3\).
4. \(\sum f_i=10\).
5. \(\bar{x}=20+10(3/10)=23\).
6. Answer: \(\boxed{23}\).

Question 5 — Median

For \(l=20,\ n=50,\ cf=18,\ f=12,\ h=10\), find the median.

1. \(n/2=25\).
2. \(\text{Median}=20+\left(\frac{25-18}{12}\right)10\).
3. \(=20+\frac{70}{12}\).
4. \(\approx25.83\).
5. Answer: \(\boxed{25.83\text{ approximately}}\).

Question 6 — Find Median Class

Total frequency is \(60\) and cumulative frequencies are \(8,17,31,46,60\). Identify the median class.

1. \(n/2=60/2=30\).
2. Find the first cumulative frequency greater than \(30\).
3. It is \(31\).
4. Therefore the class corresponding to cf \(31\) is the median class.
5. Answer: The class with \(\boxed{cf=31}\).

Question 7 — Mode

For \(l=30,\ f_0=7,\ f_1=12,\ f_2=8,\ h=10\), find the mode.

1. \(\text{Mode}=30+\left(\frac{12-7}{24-7-8}\right)10\).
2. \(=30+\frac5{9}\times10\).
3. \(\approx30+5.56=35.56\).
4. Answer: \(\boxed{35.56\text{ approximately}}\).

Question 8 — Empirical Relation

The mean is \(25\) and median is \(27\). Find the mode.

1. \(\text{Mode}=3(27)-2(25)\).
2. \(=81-50=31\).
3. Answer: \(\boxed{31}\).

Question 9 — Mean with Frequencies

Find the mean of values \(5,10,15,20\) with frequencies \(2,3,4,1\).

1. \(f_ix_i=10,30,60,20\).
2. \(\sum f_i=2+3+4+1=10\).
3. \(\sum f_ix_i=120\).
4. \(\bar{x}=120/10=12\).
5. Answer: \(\boxed{12}\).

Question 10 — Full Statistics Practice

For a grouped distribution, \(n=40,\ l=30,\ cf=14,\ f=10,\ h=10\). Also, mean \(=34\). Find the median and then the mode using the empirical relation.

1. Median \(=30+\left(\frac{20-14}{10}\right)10\).
2. Median \(=30+6=36\).
3. Use \(\text{Mode}=3(\text{Median})-2(\text{Mean})\).
4. \(=3(36)-2(34)\).
5. \(=108-68=40\).
6. Answer: Median \(=\boxed{36}\), Mode \(=\boxed{40}\).

11. Board Exam Strategy

High-Scoring Method

1. Make the table neatly. Keep class, frequency, class mark and required columns aligned.
2. Do not skip \(n=\sum f\). It is essential in median questions.
3. For median, calculate cumulative frequency carefully.
4. For mode, identify the highest frequency first.
5. Write the formula before substitution.
6. Keep calculations organised. One small arithmetic mistake can change the final answer.
7. Write the final answer clearly.

Most Important Checks

• Class mark \(=\frac{\text{lower}+\text{upper}}2\).
• Total frequency \(n=\sum f\).
• Median class is found using \(n/2\).
• Modal class has the highest frequency.
• Use the correct \(cf\), \(f\), \(f_0\), \(f_1\), \(f_2\).
• Use the correct class size \(h\).

12. Quick Revision Sheet

Concept	Formula / Key Fact
Class Mark	\(x_i=\frac{\text{lower}+\text{upper}}2\)
Direct Mean	\(\bar{x}=\frac{\sum f_ix_i}{\sum f_i}\)
Deviation	\(d_i=x_i-a\)
Assumed Mean	\(\bar{x}=a+\frac{\sum f_id_i}{\sum f_i}\)
Step Deviation	\(\bar{x}=a+h\frac{\sum f_iu_i}{\sum f_i}\)
Step Deviation \(u_i\)	\(u_i=\frac{x_i-a}{h}\)
Median	\(l+\left(\frac{n/2-cf}{f}\right)h\)
Mode	\(l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h\)
Empirical Relation	\(\text{Mode}=3\text{Median}-2\text{Mean}\)
Total Frequency	\(n=\sum f\)

30-Second Final Revision

• Mean: average value of the data.
• Class mark: midpoint of a class.
• Median: middle-position measure.
• Mode: most frequently occurring value.
• Median class: class whose cf first exceeds \(n/2\).
• Modal class: class with the highest frequency.
• Direct method: use \(f_ix_i\).
• Assumed mean: use deviations \(d_i\).
• Step deviation: use \(u_i=(x_i-a)/h\).

Common Mistakes to Avoid

• Forgetting to calculate class marks correctly.
• Using frequency instead of cumulative frequency for \(cf\).
• Choosing the wrong median class.
• Choosing a class other than the highest-frequency class as modal class.
• Mixing \(f_0,f_1,f_2\) in the mode formula.
• Using the wrong class size \(h\).
• Forgetting the \(n/2\) term in the median formula.
• Making arithmetic mistakes in \(\sum f_ix_i\).

Last-Minute Trick: For every Statistics numerical, follow Table → Required Column → Formula → Substitution → Calculation → Final Answer. This keeps the solution neat and reduces errors.

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Class 10 Mathematics • Chapter 14: Statistics

Detailed • Easy English • Board-Exam Focused • Handwritten-Style Notes

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