Standard Deviation: Complete Guide
What is Standard Deviation?
Standard deviation is a statistical measure that quantifies the amount of dispersion or variation in a set of data values. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.
Key Concept: Standard deviation represents the average distance between each data point and the mean.
Why Standard Deviation Matters
Standard deviation is important because:
- It provides a standardized measure of dispersion
- It uses the same units as the original data
- It's sensitive to outliers (which can be both an advantage and disadvantage)
- It's widely used in statistics for inference, hypothesis testing, and constructing confidence intervals
Visual Representation
In a normal distribution, approximately:
- 68% of the data falls within one standard deviation of the mean
- 95% falls within two standard deviations
- 99.7% falls within three standard deviations
Types of Standard Deviation
1. Population Standard Deviation (σ)
Used when you have data for the entire population (all possible observations).
Population Standard Deviation (σ):
σ = √[ Σ(X - μ)² / N ]
Where:
- X = each value in the population
- μ = the population mean
- N = the number of values in the population
- Σ = sum of
σ = √[ Σ(X - μ)² / N ]
Where:
- X = each value in the population
- μ = the population mean
- N = the number of values in the population
- Σ = sum of
2. Sample Standard Deviation (s)
Used when you have data from a sample (subset) of the population.
Sample Standard Deviation (s):
s = √[ Σ(x - x̄)² / (n-1) ]
Where:
- x = each value in the sample
- x̄ = the sample mean
- n = the number of values in the sample
- Σ = sum of
s = √[ Σ(x - x̄)² / (n-1) ]
Where:
- x = each value in the sample
- x̄ = the sample mean
- n = the number of values in the sample
- Σ = sum of
Important Difference: Notice that the sample standard deviation divides by (n-1) instead of N. This is called "Bessel's correction" and it helps to correct the bias in the estimation of the population standard deviation.
3. Corrected Sample Standard Deviation
Sometimes a finite population correction (FPC) is applied when the sample size is a significant portion of the population size.
Corrected Sample Standard Deviation:
s_corrected = s * √[(N-n)/(N-1)]
Where:
- s = sample standard deviation
- N = population size
- n = sample size
s_corrected = s * √[(N-n)/(N-1)]
Where:
- s = sample standard deviation
- N = population size
- n = sample size
When to Use Each Type
| Type | Use When | Symbol |
|---|---|---|
| Population | You have data for all possible observations | σ (sigma) |
| Sample | You have data from only a subset of the population | s |
| Corrected Sample | Your sample is a large proportion of the population | s_corrected |
How to Calculate Standard Deviation
Let's explore several methods for calculating standard deviation, from basic step-by-step approaches to shortcut formulas.
Method 1: Step-by-Step Calculation (Definition Formula)
- Calculate the mean (average) of the data set
- Subtract the mean from each data point to find deviations
- Square each deviation
- Sum all squared deviations
- Divide by N (population) or N-1 (sample)
- Take the square root of the result
Example: Calculate the sample standard deviation of {4, 8, 6, 5, 3, 8}
Step 1: Calculate the mean: (4+8+6+5+3+8)/6 = 34/6 = 5.67
Step 2: Find deviations from the mean:
4 - 5.67 = -1.67
8 - 5.67 = 2.33
6 - 5.67 = 0.33
5 - 5.67 = -0.67
3 - 5.67 = -2.67
8 - 5.67 = 2.33
Step 3: Square each deviation:
(-1.67)² = 2.79
(2.33)² = 5.43
(0.33)² = 0.11
(-0.67)² = 0.45
(-2.67)² = 7.13
(2.33)² = 5.43
Step 4: Sum the squared deviations:
2.79 + 5.43 + 0.11 + 0.45 + 7.13 + 5.43 = 21.34
Step 5: Divide by (n-1) for a sample:
21.34 / 5 = 4.27
Step 6: Take the square root:
√4.27 ≈ 2.07
Therefore, the sample standard deviation is approximately 2.07.
Step 1: Calculate the mean: (4+8+6+5+3+8)/6 = 34/6 = 5.67
Step 2: Find deviations from the mean:
4 - 5.67 = -1.67
8 - 5.67 = 2.33
6 - 5.67 = 0.33
5 - 5.67 = -0.67
3 - 5.67 = -2.67
8 - 5.67 = 2.33
Step 3: Square each deviation:
(-1.67)² = 2.79
(2.33)² = 5.43
(0.33)² = 0.11
(-0.67)² = 0.45
(-2.67)² = 7.13
(2.33)² = 5.43
Step 4: Sum the squared deviations:
2.79 + 5.43 + 0.11 + 0.45 + 7.13 + 5.43 = 21.34
Step 5: Divide by (n-1) for a sample:
21.34 / 5 = 4.27
Step 6: Take the square root:
√4.27 ≈ 2.07
Therefore, the sample standard deviation is approximately 2.07.
Method 2: Variance Formula (Shortcut Method)
Standard deviation is the square root of variance. This method uses algebraic simplification.
Population Variance:
σ² = [ Σ(X²) / N ] - μ²
Sample Variance:
s² = [ Σ(x²) / (n-1) ] - [ (Σx)² / n(n-1) ]
Then take the square root to get the standard deviation.
σ² = [ Σ(X²) / N ] - μ²
Sample Variance:
s² = [ Σ(x²) / (n-1) ] - [ (Σx)² / n(n-1) ]
Then take the square root to get the standard deviation.
Example: Using the same data {4, 8, 6, 5, 3, 8}
Step 1: Calculate Σx (sum of all values) = 4+8+6+5+3+8 = 34
Step 2: Calculate Σx² (sum of squared values) = 4²+8²+6²+5²+3²+8² = 16+64+36+25+9+64 = 214
Step 3: Calculate (Σx)² = 34² = 1156
Step 4: Apply the formula for sample variance:
s² = [ 214 / 5 ] - [ 1156 / (6×5) ]
s² = 42.8 - 38.53
s² = 4.27
Step 5: Calculate standard deviation:
s = √4.27 ≈ 2.07
Step 1: Calculate Σx (sum of all values) = 4+8+6+5+3+8 = 34
Step 2: Calculate Σx² (sum of squared values) = 4²+8²+6²+5²+3²+8² = 16+64+36+25+9+64 = 214
Step 3: Calculate (Σx)² = 34² = 1156
Step 4: Apply the formula for sample variance:
s² = [ 214 / 5 ] - [ 1156 / (6×5) ]
s² = 42.8 - 38.53
s² = 4.27
Step 5: Calculate standard deviation:
s = √4.27 ≈ 2.07
Method 3: Using Technology
a) Using Spreadsheets
- Excel: =STDEV.S(range) for sample; =STDEV.P(range) for population
- Google Sheets: =STDEV(range) for sample; =STDEVP(range) for population
b) Using Scientific Calculators
- Most scientific calculators have built-in functions for calculating standard deviation
- Typically labeled as "σn" or "σn-1" (or similar notation)
c) Using Statistical Software
- R: sd(data)
- Python: numpy.std(data, ddof=0) for population; numpy.std(data, ddof=1) for sample
- SPSS: Descriptive Statistics procedure
Standard Deviation Examples
Example 1: Simple Dataset
Dataset: Test scores {85, 90, 72, 95, 83}
Step 1: Calculate the mean: (85+90+72+95+83)/5 = 425/5 = 85
Step 2: Find deviations: (85-85)=0, (90-85)=5, (72-85)=-13, (95-85)=10, (83-85)=-2
Step 3: Square deviations: 0², 5², (-13)², 10², (-2)² = 0, 25, 169, 100, 4
Step 4: Sum squared deviations: 0+25+169+100+4 = 298
Step 5: Divide by (n-1): 298/4 = 74.5
Step 6: Take the square root: √74.5 ≈ 8.63
The sample standard deviation is 8.63.
Step 1: Calculate the mean: (85+90+72+95+83)/5 = 425/5 = 85
Step 2: Find deviations: (85-85)=0, (90-85)=5, (72-85)=-13, (95-85)=10, (83-85)=-2
Step 3: Square deviations: 0², 5², (-13)², 10², (-2)² = 0, 25, 169, 100, 4
Step 4: Sum squared deviations: 0+25+169+100+4 = 298
Step 5: Divide by (n-1): 298/4 = 74.5
Step 6: Take the square root: √74.5 ≈ 8.63
The sample standard deviation is 8.63.
Example 2: Comparing Datasets
Dataset A: {10, 12, 11, 13, 9, 15}
Dataset B: {2, 18, 5, 20, 14, 1}
Both datasets have the same mean of 11.67, but their spread is different.
Dataset A Standard Deviation:
Mean = 11.67
Squared deviations: (10-11.67)² + (12-11.67)² + ... = 28.84
Variance = 28.84/5 = 5.77
Standard deviation = √5.77 ≈ 2.40
Dataset B Standard Deviation:
Mean = 11.67
Squared deviations: (2-11.67)² + (18-11.67)² + ... = 415.34
Variance = 415.34/5 = 83.07
Standard deviation = √83.07 ≈ 9.11
Interpretation: Dataset B has a much higher standard deviation (9.11 vs. 2.40), indicating that its values are spread out more widely from the mean.
Dataset B: {2, 18, 5, 20, 14, 1}
Both datasets have the same mean of 11.67, but their spread is different.
Dataset A Standard Deviation:
Mean = 11.67
Squared deviations: (10-11.67)² + (12-11.67)² + ... = 28.84
Variance = 28.84/5 = 5.77
Standard deviation = √5.77 ≈ 2.40
Dataset B Standard Deviation:
Mean = 11.67
Squared deviations: (2-11.67)² + (18-11.67)² + ... = 415.34
Variance = 415.34/5 = 83.07
Standard deviation = √83.07 ≈ 9.11
Interpretation: Dataset B has a much higher standard deviation (9.11 vs. 2.40), indicating that its values are spread out more widely from the mean.
Example 3: Continuous Data
Dataset: Heights of a sample of 7 people (in cm): {168, 172, 165, 175, 178, 169, 173}
Step 1: Calculate the mean: (168+172+165+175+178+169+173)/7 = 1200/7 ≈ 171.43
Step 2: Find deviations: (168-171.43), (172-171.43), etc.
Step 3: Square deviations: (-3.43)², (0.57)², etc.
Step 4: Sum squared deviations = 127.43
Step 5: Divide by (n-1): 127.43/6 ≈ 21.24
Step 6: Take the square root: √21.24 ≈ 4.61
Therefore, the standard deviation of heights is 4.61 cm.
Step 1: Calculate the mean: (168+172+165+175+178+169+173)/7 = 1200/7 ≈ 171.43
Step 2: Find deviations: (168-171.43), (172-171.43), etc.
Step 3: Square deviations: (-3.43)², (0.57)², etc.
Step 4: Sum squared deviations = 127.43
Step 5: Divide by (n-1): 127.43/6 ≈ 21.24
Step 6: Take the square root: √21.24 ≈ 4.61
Therefore, the standard deviation of heights is 4.61 cm.
Applications of Standard Deviation
1. Quality Control
Example: A manufacturing process produces bolts with a target diameter of 10mm. Quality control engineers establish control limits at ±3 standard deviations. If the standard deviation is 0.05mm, any bolt measuring outside 9.85mm to 10.15mm would be rejected.
2. Finance and Investment
Example: A stock with an average annual return of 8% and a standard deviation of 15% is more volatile (riskier) than a stock with the same return but a standard deviation of 5%. In a normal distribution, this means the first stock's returns will fall between -7% and 23% about 68% of the time, while the second stock's returns will fall between 3% and 13% with the same probability.
3. Weather Forecasting
Example: Meteorologists might report that the average July temperature in New York is 84°F with a standard deviation of 4°F. This tells us that most July days (about 68%) have temperatures between 80°F and 88°F.
4. Educational Assessment
Example: Standardized tests like the SAT are designed with a mean of 1000 and a standard deviation of 200. This means that about 68% of test-takers score between 800 and 1200, and about 95% score between 600 and 1400.
5. Biological Research
Example: When testing a new drug, researchers might report that it lowers cholesterol by an average of 15% with a standard deviation of 3%. This gives other scientists information about both the effectiveness and the consistency of the drug's effects.
6. Z-Scores and Standardization
Z-scores represent how many standard deviations a data point is from the mean.
Z-Score:
Z = (X - μ) / σ
Where:
- X = the data point
- μ = the mean
- σ = the standard deviation
Z = (X - μ) / σ
Where:
- X = the data point
- μ = the mean
- σ = the standard deviation
Example: In a class where the mean test score is 75 with a standard deviation of 8, a student who scored 91 would have a z-score of (91-75)/8 = 2. This means they scored 2 standard deviations above the mean, placing them among the top 2.3% of the class (assuming a normal distribution).
Standard Deviation Calculator
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Standard Deviation Quiz
1. What does a standard deviation of 0 indicate?
a) All data values are identical (no variation)
b) The data follows a normal distribution
c) There are no outliers in the data
d) The mean of the data is 0
2. In a standard normal distribution, approximately what percentage of data falls within 1 standard deviation of the mean?
a) 50%
b) 68%
c) 95%
d) 99.7%
3. The standard deviation of a data set is 5. If every value in the data set is multiplied by 2, what happens to the standard deviation?
a) It becomes 10
b) It becomes 7
c) It remains 5
d) It becomes 25
4. Which formula is used to calculate the sample standard deviation?
a) s = √[ Σ(x - x̄)² / n ]
b) s = √[ Σ(x - x̄)² / (n-1) ]
c) s = Σ|x - x̄| / n
d) s = √[ Σx² / n ]
5. Calculate the standard deviation of the data set {2, 4, 6, 8, 10}.
a) 2.83
b) 3.16
c) 8
d) 10
6. If a data point has a z-score of 2, this means it is:
a) Equal to the mean
b) 2 units above the mean
c) 2 standard deviations above the mean
d) Twice as large as the mean
7. When should you use population standard deviation instead of sample standard deviation?
a) When you have data for every member of the population
b) When your sample size is very large
c) When the data follows a normal distribution
d) When you're working with continuous data
8. Which of the following statements is true about standard deviation?
a) It can be negative
b) It is always less than or equal to the range
c) It is sensitive to outliers
d) It is always equal to the square root of the mean
9. Which of these data sets has the smallest standard deviation?
a) {1, 1, 1, 99}
b) {10, 20, 30, 40}
c) {25, 25, 25, 25}
d) {0, 33, 67, 100}
10. The main difference between population and sample standard deviation formulas is:
a) Sample uses (n-1) in the denominator instead of n
b) Population uses the square of the deviations, sample does not
c) Sample requires at least 30 data points
d) Population requires a normal distribution