The Standard Deviation Formula measures how spread out data values are from their mean, expressed as \ ( \sigma = \sqrt{\frac{\sum (x_i – \bar{x})^2}{N}} \) for a population. It is a core concept in NCERT Class 11 Mathematics (Chapter 15: Statistics) and appears regularly in CBSE board exams, JEE Main, and even NEET biology data-interpretation questions. This article covers every form of the standard deviation formula, a complete statistics formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Standard Deviation Formulas at a Glance
Quick reference for the most important standard deviation expressions.
- Population SD: \( \sigma = \sqrt{\dfrac{\sum_{i=1}^{N}(x_i – \bar{x})^2}{N}} \)
- Sample SD: \( s = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i – \bar{x})^2}{n-1}} \)
- Shortcut (direct) method: \( \sigma = \sqrt{\dfrac{\sum x_i^2}{N} – \bar{x}^2} \)
- For frequency distribution: \( \sigma = \sqrt{\dfrac{\sum f_i(x_i – \bar{x})^2}{\sum f_i}} \)
- Coefficient of Variation: \( CV = \dfrac{\sigma}{\bar{x}} \times 100\% \)
- Variance: \( \sigma^2 = \dfrac{\sum (x_i – \bar{x})^2}{N} \)
- Step-deviation method: \( \sigma = h \cdot \sqrt{\dfrac{\sum f_i u_i^2}{\sum f_i} – \left(\dfrac{\sum f_i u_i}{\sum f_i}\right)^2} \)
What is the Standard Deviation Formula?
The Standard Deviation Formula is a statistical tool that quantifies the amount of variation or dispersion in a set of data values. A low standard deviation means data points are clustered closely around the mean. A high standard deviation means data points are spread widely.
In NCERT Class 11 Mathematics, Chapter 15 (Statistics), standard deviation is introduced as the most reliable measure of dispersion. It improves upon the mean deviation by squaring the deviations, which removes negative signs and penalises larger deviations more heavily.
The symbol for population standard deviation is \( \sigma \) (sigma), while sample standard deviation uses \( s \). Variance, denoted \( \sigma^2 \), is simply the square of the standard deviation. Together, these measures help students analyse real-world data — from exam scores to stock prices — with mathematical precision. Understanding the Standard Deviation Formula is essential for scoring well in CBSE Statistics and for data-interpretation problems in JEE Main and NEET.
Standard Deviation Formula — Expression and Variables
1. Population Standard Deviation
\[ \sigma = \sqrt{\frac{\sum_{i=1}^{N}(x_i – \bar{x})^2}{N}} \]
2. Sample Standard Deviation
\[ s = \sqrt{\frac{\sum_{i=1}^{n}(x_i – \bar{x})^2}{n – 1}} \]
3. Shortcut / Direct Method
\[ \sigma = \sqrt{\frac{\sum x_i^2}{N} – \left(\frac{\sum x_i}{N}\right)^2} \]
4. Frequency Distribution Formula
\[ \sigma = \sqrt{\frac{\sum f_i (x_i – \bar{x})^2}{\sum f_i}} \]
| Symbol | Quantity | SI Unit / Note |
|---|---|---|
| \( \sigma \) | Population standard deviation | Same unit as data |
| \( s \) | Sample standard deviation | Same unit as data |
| \( x_i \) | Each individual data value | Same unit as data |
| \( \bar{x} \) | Arithmetic mean of the data | Same unit as data |
| \( N \) | Total number of data points (population) | Dimensionless |
| \( n \) | Number of data points in sample | Dimensionless |
| \( f_i \) | Frequency of the \( i \)-th class | Dimensionless |
| \( \sigma^2 \) | Variance | Square of data unit |
| \( h \) | Class width (step-deviation method) | Same unit as data |
| \( u_i \) | Step deviation: \( u_i = (x_i – A)/h \) | Dimensionless |
Derivation of the Population Standard Deviation Formula
The derivation begins with the concept of variance. First, calculate the mean: \( \bar{x} = \dfrac{\sum x_i}{N} \). Next, find each deviation from the mean: \( (x_i – \bar{x}) \). Square every deviation to eliminate negatives: \( (x_i – \bar{x})^2 \). Sum all squared deviations and divide by \( N \) to get the variance: \( \sigma^2 = \dfrac{\sum (x_i – \bar{x})^2}{N} \). Finally, take the positive square root to return to the original unit of measurement. This gives the Standard Deviation Formula: \( \sigma = \sqrt{\sigma^2} \). The shortcut formula is derived by expanding the squared term and simplifying, yielding \( \sigma = \sqrt{\dfrac{\sum x_i^2}{N} – \bar{x}^2} \), which reduces computation significantly.
Complete Statistics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Arithmetic Mean | \( \bar{x} = \dfrac{\sum x_i}{N} \) | \( x_i \) = data values, \( N \) = count | Same as data | Class 11, Ch 15 |
| Population Variance | \( \sigma^2 = \dfrac{\sum (x_i – \bar{x})^2}{N} \) | \( x_i \) = data, \( \bar{x} \) = mean, \( N \) = count | Square of data unit | Class 11, Ch 15 |
| Population Standard Deviation | \( \sigma = \sqrt{\dfrac{\sum (x_i – \bar{x})^2}{N}} \) | \( x_i \) = data, \( \bar{x} \) = mean, \( N \) = count | Same as data | Class 11, Ch 15 |
| Sample Standard Deviation | \( s = \sqrt{\dfrac{\sum (x_i – \bar{x})^2}{n-1}} \) | \( n \) = sample size | Same as data | Class 11, Ch 15 |
| Shortcut SD Formula | \( \sigma = \sqrt{\dfrac{\sum x_i^2}{N} – \bar{x}^2} \) | \( x_i \) = data, \( \bar{x} \) = mean | Same as data | Class 11, Ch 15 |
| SD for Frequency Distribution | \( \sigma = \sqrt{\dfrac{\sum f_i(x_i-\bar{x})^2}{\sum f_i}} \) | \( f_i \) = frequency, \( x_i \) = mid-value | Same as data | Class 11, Ch 15 |
| Step-Deviation Method SD | \( \sigma = h\sqrt{\dfrac{\sum f_i u_i^2}{\sum f_i} – \left(\dfrac{\sum f_i u_i}{\sum f_i}\right)^2} \) | \( h \) = class width, \( u_i = (x_i-A)/h \) | Same as data | Class 11, Ch 15 |
| Coefficient of Variation | \( CV = \dfrac{\sigma}{\bar{x}} \times 100\% \) | \( \sigma \) = SD, \( \bar{x} \) = mean | Percentage (%) | Class 11, Ch 15 |
| Mean Deviation about Mean | \( MD = \dfrac{\sum |x_i – \bar{x}|}{N} \) | \( x_i \) = data, \( \bar{x} \) = mean | Same as data | Class 11, Ch 15 |
| Range | \( R = x_{\max} – x_{\min} \) | Max and min data values | Same as data | Class 11, Ch 15 |
Standard Deviation Formula — Solved Examples
Example 1 (Class 9-10 Level): SD of a Small Data Set
Problem: Find the standard deviation of the data set: 4, 8, 6, 5, 3, 2, 8, 9, 2, 5.
Given: Data values: 4, 8, 6, 5, 3, 2, 8, 9, 2, 5. Total number of values \( N = 10 \).
Step 1: Calculate the mean.
\( \bar{x} = \dfrac{4+8+6+5+3+2+8+9+2+5}{10} = \dfrac{52}{10} = 5.2 \)
Step 2: Find each deviation \( (x_i – \bar{x}) \) and square it.
| \( x_i \) | \( x_i – \bar{x} \) | \( (x_i – \bar{x})^2 \) |
|---|---|---|
| 4 | -1.2 | 1.44 |
| 8 | 2.8 | 7.84 |
| 6 | 0.8 | 0.64 |
| 5 | -0.2 | 0.04 |
| 3 | -2.2 | 4.84 |
| 2 | -3.2 | 10.24 |
| 8 | 2.8 | 7.84 |
| 9 | 3.8 | 14.44 |
| 2 | -3.2 | 10.24 |
| 5 | -0.2 | 0.04 |
Step 3: Sum the squared deviations.
\( \sum (x_i – \bar{x})^2 = 1.44 + 7.84 + 0.64 + 0.04 + 4.84 + 10.24 + 7.84 + 14.44 + 10.24 + 0.04 = 57.6 \)
Step 4: Apply the Standard Deviation Formula.
\( \sigma = \sqrt{\dfrac{57.6}{10}} = \sqrt{5.76} = 2.4 \)
Answer
Standard Deviation \( \sigma = 2.4 \)
Example 2 (Class 11-12 Level): SD Using Frequency Distribution
Problem: The marks scored by 50 students in a test are given in the frequency table below. Find the standard deviation.
| Marks (\( x_i \)) | Frequency (\( f_i \)) |
|---|---|
| 10 | 5 |
| 20 | 8 |
| 30 | 15 |
| 40 | 12 |
| 50 | 10 |
Given: \( \sum f_i = 50 \)
Step 1: Calculate the mean.
\( \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{(5)(10)+(8)(20)+(15)(30)+(12)(40)+(10)(50)}{50} \)
\( = \dfrac{50+160+450+480+500}{50} = \dfrac{1640}{50} = 32.8 \)
Step 2: Compute \( f_i(x_i – \bar{x})^2 \) for each class.
| \( x_i \) | \( f_i \) | \( x_i – \bar{x} \) | \( (x_i-\bar{x})^2 \) | \( f_i(x_i-\bar{x})^2 \) |
|---|---|---|---|---|
| 10 | 5 | -22.8 | 519.84 | 2599.2 |
| 20 | 8 | -12.8 | 163.84 | 1310.72 |
| 30 | 15 | -2.8 | 7.84 | 117.6 |
| 40 | 12 | 7.2 | 51.84 | 622.08 |
| 50 | 10 | 17.2 | 295.84 | 2958.4 |
Step 3: Sum the last column.
\( \sum f_i(x_i – \bar{x})^2 = 2599.2 + 1310.72 + 117.6 + 622.08 + 2958.4 = 7608 \)
Step 4: Apply the frequency distribution formula.
\( \sigma = \sqrt{\dfrac{7608}{50}} = \sqrt{152.16} \approx 12.34 \)
Answer
Standard Deviation \( \sigma \approx 12.34 \) marks
Example 3 (JEE/NEET Level): Coefficient of Variation and Comparison
Problem: Two batsmen A and B scored runs in 6 innings as follows. Determine which batsman is more consistent using the Coefficient of Variation.
Batsman A: 25, 85, 40, 80, 120, 10
Batsman B: 55, 70, 53, 58, 60, 64
Given: \( N = 6 \) for both batsmen.
Step 1: For Batsman A
\( \bar{x}_A = \dfrac{25+85+40+80+120+10}{6} = \dfrac{360}{6} = 60 \)
Deviations: -35, 25, -20, 20, 60, -50
Squared deviations: 1225, 625, 400, 400, 3600, 2500
\( \sum(x_i – \bar{x})^2 = 8750 \)
\( \sigma_A = \sqrt{\dfrac{8750}{6}} = \sqrt{1458.33} \approx 38.19 \)
\( CV_A = \dfrac{38.19}{60} \times 100 \approx 63.65\% \)
Step 2: For Batsman B
\( \bar{x}_B = \dfrac{55+70+53+58+60+64}{6} = \dfrac{360}{6} = 60 \)
Deviations: -5, 10, -7, -2, 0, 4
Squared deviations: 25, 100, 49, 4, 0, 16
\( \sum(x_i – \bar{x})^2 = 194 \)
\( \sigma_B = \sqrt{\dfrac{194}{6}} = \sqrt{32.33} \approx 5.69 \)
\( CV_B = \dfrac{5.69}{60} \times 100 \approx 9.48\% \)
Step 3: Compare. Both batsmen have the same mean (60 runs). However, \( CV_B = 9.48\% \ll CV_A = 63.65\% \). A lower CV indicates greater consistency.
Answer
Batsman B is significantly more consistent than Batsman A, because Batsman B has a much lower Coefficient of Variation (9.48% vs 63.65%).
CBSE Exam Tips 2025-26
- Memorise all four forms: We recommend learning the population, sample, shortcut, and frequency distribution versions separately. CBSE 2025-26 papers often test the frequency distribution form in 4-mark questions.
- Use the shortcut formula for large data: When data values are large, the shortcut formula \( \sigma = \sqrt{\dfrac{\sum x_i^2}{N} – \bar{x}^2} \) saves significant time. Always verify your mean first.
- Show every step: CBSE awards step marks. Write the formula, substitute values, and simplify in clearly numbered steps. Never skip the variance step.
- Coefficient of Variation questions: CBSE Class 11 frequently asks comparison problems. Our experts suggest practising at least 5 CV comparison problems before the exam.
- Check units: Standard deviation always has the same unit as the original data. Variance has squared units. Mentioning units in your answer earns full marks.
- Step-deviation method for grouped data: For continuous frequency distributions, the step-deviation method is the fastest approach. Choose the assumed mean \( A \) as the middle class mark to minimise calculation errors.
Common Mistakes to Avoid
- Forgetting to square root the variance: Many students calculate variance correctly but forget to take the square root. Variance is \( \sigma^2 \); standard deviation is \( \sigma \). Always complete the final step.
- Using \( n \) instead of \( n-1 \) for sample data: When the problem says “sample,” use Bessel’s correction (divide by \( n-1 \)). Dividing by \( n \) gives a biased estimate for samples. CBSE Class 11 generally uses population SD, but JEE problems may specify sample SD.
- Incorrect mean calculation: A wrong mean cascades into every subsequent step. Double-check \( \bar{x} \) before proceeding. This is the single most common source of error in board exams.
- Confusing mean deviation with standard deviation: Mean deviation uses absolute values \( |x_i – \bar{x}| \). Standard deviation squares the deviations. They are different measures and yield different numerical results.
- Ignoring the class mid-point for grouped data: In a frequency distribution, \( x_i \) must be the class mid-point, not the class boundary. Using the upper or lower boundary is a common error that changes the answer entirely.
JEE/NEET Application of Standard Deviation Formula
In our experience, JEE aspirants encounter the Standard Deviation Formula in at least one question per JEE Main session, typically in the “Statistics” chapter of Class 11 Mathematics. NEET aspirants encounter it in biology data-interpretation and experimental design questions.
JEE Main Application Patterns
- Pattern 1 — Change in SD when data is transformed: JEE frequently asks: “If each observation is multiplied by a constant \( k \), what happens to the standard deviation?” The answer is that SD also gets multiplied by \( |k| \). Adding a constant to every value does NOT change the SD. This property is tested almost every year.
- Pattern 2 — Variance of combined data: JEE Advanced problems combine two data sets and ask for the combined variance. The formula involves both the individual variances and the difference between individual means and the combined mean. Practising NCERT miscellaneous examples thoroughly prepares you for this type.
- Pattern 3 — Coefficient of Variation for comparison: JEE Main often presents two data sets with the same mean but different SDs, or the same SD but different means, and asks which is more variable. Always compute CV and compare. A lower CV means less variability and greater consistency.
NEET Application Patterns
NEET Biology and Biotechnology sections include experimental data analysis. Understanding standard deviation helps interpret error bars in graphs and evaluate experimental reliability. A small SD in a biology experiment indicates reproducibility — a key concept in research methodology questions in NEET 2025-26.
Our experts suggest solving all NCERT Class 11 Chapter 15 examples and exercises. Then practise 10-15 previous year JEE Main Statistics questions. This builds the speed and accuracy needed for competitive exams.
FAQs on Standard Deviation Formula
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Strengthen your Mathematics preparation with these related resources on ncertbooks.net. Study the Vertex Formula to master quadratic functions alongside statistics. Review the complete Algebra Formulas hub for a structured overview of all Class 11-12 algebraic identities. You can also explore the Sine Rule Formula to build your trigonometry foundation. For the official NCERT syllabus reference, visit the NCERT official website.