01 · Explore
Introduction to Dispersion
While measures of central tendency like mean and median provide a representative value for a dataset, they do not describe how the data is spread out.
Consider two batsmen, A and B, who both have a mean score of 53. Batsman A's scores range from 0 to 117, showing high variability. Batsman B's scores range from 46 to 60, showing consistency. Although their 'center' is the same, their performance profiles are vastly different.
Statistics uses 'measures of dispersion' to quantify this scatter. A single number can describe how much data points are bunched around or scattered away from the central tendency. In this chapter, we explore Range, Mean Deviation, Variance, and Standard Deviation.
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Why is mean alone insufficient to describe a dataset?
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02 · Explore
Range and Mean Deviation
The simplest measure of dispersion is the Range, while Mean Deviation provides a more detailed look at individual distances from a central value.
Range is defined as the difference between the maximum and minimum values in a series: Range = Maximum value – Minimum value. While easy to calculate, it only uses two data points and ignores the distribution of the rest of the data.
Mean Deviation (M.D.) improves on this by considering the absolute difference (distance) of every observation from a central value 'a' (usually the mean or median). We use absolute values because the sum of simple deviations from the mean is always zero, which would be useless for measuring spread.
| Measure | Formula (Ungrouped) | Description |
|---|---|---|
| Range | Max - Min | Difference between extremes. |
| M.D. about Mean | 1/n * Σ |xi - x| | Mean of absolute distances from the mean. |
| M.D. about Median | 1/n * Σ |xi - M| | Mean of absolute distances from the median. |
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Why do we take absolute values in Mean Deviation?
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03 · Explore
Worked Example: M.D. for Ungrouped Data
Let's calculate the mean deviation about the mean for a specific set of observations.
To find the M.D. about the mean, we first calculate the arithmetic mean, then find the absolute difference of each observation from that mean, and finally find the average of those differences.
Calculating M.D. about Mean
Data: 6, 7, 10, 12, 13, 4, 8, 12
Step 1: Find Mean (x). x = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9. Step 2: Find absolute deviations |xi - 9|: |6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3. Step 3: Sum of deviations = 3+2+1+3+4+5+1+3 = 22. Step 4: M.D.(x) = 22/8 = 2.75.
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Find the M.D. about the mean for: 3, 5, 7, 9, 11.
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04 · Explore
Mean Deviation for Discrete Frequency Distribution
When data points occur with specific frequencies, the calculation must account for the weight of each observation.
In a discrete frequency distribution, we have values xi with frequencies fi. The total number of observations N is the sum of all frequencies (Σfi).
The formula for Mean Deviation about the mean becomes: M.D.(x) = (1/N) * Σ fi |xi - x|. Similarly, for the median: M.D.(M) = (1/N) * Σ fi |xi - M|.
Process for M.D. (Discrete Frequency)
- 1
Calculate Central Value
Find the mean (Σfixi / N) or the median (using cumulative frequency).
- 2
Find Deviations
Calculate |xi - a| for every observation.
- 3
Weight Deviations
Multiply each absolute deviation by its frequency: fi |xi - a|.
- 4
Final Average
Sum these products and divide by the total frequency N.
Sequential steps to find Mean Deviation in a frequency-based dataset.
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How is the median found in a discrete frequency distribution?
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05 · Explore
Mean Deviation for Continuous Data
Continuous data involves class intervals. We treat these by using the mid-point of each class.
For continuous frequency distributions (e.g., 10-20, 20-30), we assume the frequency of each class is centered at its mid-point (xi). Once mid-points are determined, the calculation follows the same logic as discrete frequency distributions.
To find the median in continuous data, we use the formula: Median = l + [ (N/2 - C) / f ] * h, where l is the lower limit of the median class, f is its frequency, h is the class width, and C is the cumulative frequency of the preceding class.
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What is the mid-point of class 30-40?
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06 · Explore
Variance and Standard Deviation
Mean deviation has limitations, such as ignoring signs via absolute values, which makes it difficult to use in higher algebra. Variance and Standard Deviation solve this by using squares.
Variance (σ²) is the mean of the squares of the deviations from the mean. Squaring the deviations ensures all values are non-negative and gives more weight to larger outliers. The formula for n observations is: σ² = (1/n) * Σ (xi - x)².
Standard Deviation (σ) is the positive square root of the variance. It is preferred because it is expressed in the same units as the original data (unlike variance, which is in squared units). σ = √[ (1/n) * Σ (xi - x)² ].
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If the variance of a set is 49, what is the standard deviation?
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What does a variance of zero imply?
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07 · Explore
Variance in Frequency Distributions
Calculating variance for grouped data involves weighting the squared deviations by their respective frequencies.
For a discrete frequency distribution, Variance σ² = (1/N) * Σ fi(xi - x)². For continuous data, we again use mid-points (xi) for the classes.
A useful alternative formula for variance that often simplifies calculation is: σ² = (1/N) * Σ fixi² - (x̄)². This avoids calculating individual deviations if the mean is not an integer.
Standard Deviation for Discrete Data (Original Example)
xi: 4, 8, 11; fi: 3, 5, 2. N=10.
1. Mean x = (4*3 + 8*5 + 11*2)/10 = (12+40+22)/10 = 7.4. 2. Squared deviations (xi-x)²: (4-7.4)²=11.56, (8-7.4)²=0.36, (11-7.4)²=12.96. 3. Weighted: 3*11.56=34.68, 5*0.36=1.8, 2*12.96=25.92. 4. Variance σ² = (34.68+1.8+25.92)/10 = 62.4/10 = 6.24. 5. σ = √6.24 ≈ 2.498.
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True or False: Standard deviation can be negative.
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08 · Explore
Shortcut Method and Properties
When dealing with large values or complex class intervals, the shortcut (step-deviation) method reduces arithmetic effort.
By defining a new variable yi = (xi - A) / h, where A is an assumed mean and h is the class width, we can find the variance of x using the variance of y. The relationship is: σx² = h² * σy².
Important Property: If a constant 'k' is added to or subtracted from every observation, the variance remains unchanged. However, if every observation is multiplied by 'k', the new variance becomes k² times the original variance, and the new standard deviation becomes |k| times the original.
New Variance = k² * Old Variance; New σ = |k| * Old σ.
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If the variance of 10 observations is 5 and each is multiplied by 3, what is the new variance?
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Does adding 10 to every value in a dataset change its standard deviation?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 23, 24, 26, 28.