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Statistics - Analysis of Frequency Distributions

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The primary objective of analyzing frequency distributions is to compare the variability, consistency, or stability of two or more data sets. This is essential when data sets have different units of measurement or significantly different means.

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The Coefficient of Variation (C.V.) is the most important tool for this analysis. It is a relative measure of dispersion, expressed as a percentage, which allows for a fair comparison between distributions regardless of their scales. Visually, a higher C.V. corresponds to a 'flatter' frequency curve where data points are widely scattered from the center.

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A distribution with a smaller Coefficient of Variation is considered more consistent, more stable, or more uniform. Conversely, a distribution with a larger C.V. is considered more variable, less stable, or more dispersed. In a graph, the more 'consistent' series will appear as a steep, narrow peak concentrated around the mean value.

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When comparing two frequency distributions with the same mean, the distribution with the smaller standard deviation (σ\sigma) is more consistent. In this case, the C.V. is directly proportional to the standard deviation, so the comparison of σ\sigma alone is sufficient to determine stability.

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Visually representing variability can be done using frequency polygons or curves. If you plot two distributions on the same set of axes, the one that looks like a tall, thin mountain has low variability (low C.V.), while the one that looks like a low, wide plateau has high variability (high C.V.).

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To calculate C.V. for a grouped frequency distribution, one must first find the arithmetic mean (xˉ\bar{x}) and the standard deviation (σ\sigma). For efficiency, the step-deviation method is often used, where di=xi−Ahd_i = \frac{x_i - A}{h} is calculated to simplify the numbers before applying the standard deviation formula.

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The analysis of frequency distributions is widely applied in finance to compare the risk (volatility) of different stocks and in sports to compare the consistency of performance between different players over a season.

📐Formulae

Arithmetic Mean (xˉ)=∑fixiN\text{Arithmetic Mean } (\bar{x}) = \frac{\sum f_i x_i}{N}

Arithmetic Mean (Step-Deviation):xˉ=A+(∑fidiN)×h\text{Arithmetic Mean (Step-Deviation)}: \bar{x} = A + \left( \frac{\sum f_i d_i}{N} \right) \times h

Standard Deviation (σ)=∑fi(xi−xˉ)2N\text{Standard Deviation } (\sigma) = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{N}}

Standard Deviation (Shortcut Method):σ=hNN∑fidi2−(∑fidi)2\text{Standard Deviation (Shortcut Method)}: \sigma = \frac{h}{N} \sqrt{N \sum f_i d_i^2 - (\sum f_i d_i)^2}

Coefficient of Variation (C.V.)=σxˉ×100\text{Coefficient of Variation (C.V.)} = \frac{\sigma}{\bar{x}} \times 100

💡Examples

Problem 1:

The following data shows the mean and standard deviation of marks obtained by two groups of students in a Mathematics test: Group A (Mean = 4545, S.D. = 99) and Group B (Mean = 5050, S.D. = 88). Which group shows more consistency in their marks?

Solution:

Step 1: Calculate the Coefficient of Variation (C.V.) for Group A. \ C.V.(A)=σAxˉA×100=945×100=20%C.V.(A) = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{9}{45} \times 100 = 20\%. \ Step 2: Calculate the Coefficient of Variation (C.V.) for Group B. \ C.V.(B)=σBxˉB×100=850×100=16%C.V.(B) = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{8}{50} \times 100 = 16\%. \ Step 3: Compare the C.V. values. Since 16%<20%16\% < 20\%, Group B has a lower C.V.

Explanation:

Consistency is determined by a lower Coefficient of Variation. Even though Group B has a higher mean, its relative dispersion (variability) is lower than Group A, making Group B more consistent.

Problem 2:

Two plants, X and Y, produce bulbs. The average life of bulbs from Plant X is 12001200 hours with a standard deviation of 120120 hours. For Plant Y, the average life is 15001500 hours with a standard deviation of 180180 hours. Which plant produces bulbs with greater relative variation?

Solution:

Step 1: Find C.V. for Plant X. \ C.V.(X)=1201200×100=10%C.V.(X) = \frac{120}{1200} \times 100 = 10\%. \ Step 2: Find C.V. for Plant Y. \ C.V.(Y)=1801500×100=12%C.V.(Y) = \frac{180}{1500} \times 100 = 12\%. \ Step 3: Compare the results. 12%>10%12\% > 10\%.

Explanation:

To find 'greater relative variation', we look for the higher C.V. Although Plant Y bulbs last longer on average, their lifespan is more variable relative to their mean compared to Plant X.