CBSE Class 11 Maths: Statistics (Measures of Dispersion)

Understanding Measures of Dispersion

Measures of central tendency (like mean, median, mode) give us a single value that represents the typical or central value of a dataset. However, this single value alone is often insufficient to fully describe a distribution. Two datasets can have the same mean but vastly different characteristics in terms of how spread out their values are. For instance, consider scores (10, 50, 90) and (48, 50, 52). Both have a mean of 50, but the first set is much more dispersed than the second.

Dispersion refers to the degree to which numerical data is likely to vary or be spread around an average value. It quantifies how spread out the values in a dataset are. A low dispersion indicates that the data points tend to be clustered closely around the mean, while a high dispersion indicates that the data points are spread out over a wider range of values. Understanding dispersion is vital for making informed decisions, as it provides insight into the reliability and consistency of data. Without knowing the dispersion, conclusions based solely on central tendency can be misleading. We will explore several key measures of dispersion including Mean Deviation, Variance, and Standard Deviation, each offering a unique perspective on data spread.

Key Measures of Dispersion

Range

The simplest measure of dispersion, defined as the difference between the maximum and minimum values in a dataset. While easy to calculate, it is highly affected by extreme values and doesn't consider the spread of intermediate values.

Quartile Deviation (Semi-Interquartile Range)

It is half the difference between the third quartile (Q3) and the first quartile (Q1). It measures the spread of the middle 50% of the data, making it less sensitive to extreme values than the range.

Mean Deviation (MD)

The arithmetic mean of the absolute deviations of the observations from their mean or median. It provides a measure of the average distance of each data point from the central value. It's calculated as $\frac{1}{N} \sum |x_i - A|$, where 'A' is either the mean or median.

Variance ($\sigma^2$)

The average of the squared differences from the mean. It quantifies the spread of data by giving more weight to larger deviations. For ungrouped data, it's $\frac{1}{N} \sum (x_i - \bar{x})^2$. For grouped data, it's $\frac{1}{N} \sum f_i (x_i - \bar{x})^2$.

Standard Deviation ($\sigma$)

The square root of the variance. It is the most commonly used measure of dispersion because it is expressed in the same units as the data, making it easier to interpret. A small standard deviation indicates data points are close to the mean, while a large standard deviation indicates data points are spread out.

Step-by-Step Calculation of Mean Deviation

1. Calculating Mean Deviation for Ungrouped Data (about Mean) — 1. Calculate the Mean ($\bar{x}$): Sum all observations and divide by the total number of observations ($N$). 2. Find Deviations: Calculate the absolute difference between each observation ($x_i$) and the mean: $|x_i - \bar{x}|$. 3. Sum Absolute Deviations: Add all the absolute deviations: $\sum |x_i - \bar{x}|$. 4. Divide by N: Mean Deviation = $\frac{\sum |x_i - \bar{x}|}{N}$.
2. Calculating Mean Deviation for Grouped Data (about Mean) — 1. Calculate the Mean ($\bar{x}$): For each class interval, find the mid-point ($x_i$). Then, $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. 2. Find Deviations: Calculate the absolute difference between each mid-point ($x_i$) and the mean: $|x_i - \bar{x}|$. 3. Multiply by Frequency: Multiply each absolute deviation by its corresponding frequency ($f_i$): $f_i |x_i - \bar{x}|$. 4. Sum Weighted Deviations: Add all the products from step 3: $\sum f_i |x_i - \bar{x}|$. 5. Divide by $\sum f_i$: Mean Deviation = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$.

Worked Examples: Mean Deviation, Variance & Standard Deviation

• Example 1: Calculate the Mean Deviation about the Mean for the data: 6, 7, 10, 12, 13, 4, 8, 12. Step 1: Calculate the Mean ($\bar{x}$) $\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$ Step 2: Find absolute deviations from the mean $|x_i - \bar{x}|$: $|6-9|=3$ $|7-9|=2$ $|10-9|=1$ $|12-9|=3$ $|13-9|=4$ $|4-9|=5$ $|8-9|=1$ $|12-9|=3$ Step 3: Sum the absolute deviations $\sum |x_i - \bar{x}| = 3+2+1+3+4+5+1+3 = 22$ Step 4: Calculate Mean Deviation Mean Deviation (MD) = $\frac{\sum |x_i - \bar{x}|}{N} = \frac{22}{8} = 2.75$ Final Answer: The Mean Deviation about the Mean is 2.75.
• Example 2: For the following distribution, calculate the Variance and Standard Deviation. | Class Interval | Frequency ($f_i$) | |---|---| | 0-10 | 2 | | 10-20 | 3 | | 20-30 | 5 | | 30-40 | 4 | | 40-50 | 1 | Step 1: Find the mid-points ($x_i$) of each class interval and calculate $\sum f_i$ | Class Interval | $f_i$ | $x_i$ | $f_i x_i$ | |---|---|---|---| | 0-10 | 2 | 5 | 10 | | 10-20 | 3 | 15 | 45 | | 20-30 | 5 | 25 | 125 | | 30-40 | 4 | 35 | 140 | | 40-50 | 1 | 45 | 45 | | Total | $\sum f_i = 15$ | | $\sum f_i x_i = 365$ | Step 2: Calculate the Mean ($\bar{x}$) $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{365}{15} \approx 24.33$ Step 3: Calculate $(x_i - \bar{x})$, $(x_i - \bar{x})^2$, and $f_i(x_i - \bar{x})^2$ | Class | $f_i$ | $x_i$ | $x_i - \bar{x}$ | $(x_i - \bar{x})^2$ | $f_i (x_i - \bar{x})^2$ | |---|---|---|---|---|---| | 0-10 | 2 | 5 | -19.33 | 373.65 | 747.30 | | 10-20 | 3 | 15 | -9.33 | 87.05 | 261.15 | | 20-30 | 5 | 25 | 0.67 | 0.45 | 2.25 | | 30-40 | 4 | 35 | 10.67 | 113.85 | 455.40 | | 40-50 | 1 | 45 | 20.67 | 427.25 | 427.25 | | Total | 15 | | | | $\sum f_i (x_i - \bar{x})^2 \approx 1893.35$ | Step 4: Calculate Variance ($\sigma^2$) $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{1893.35}{15} \approx 126.22$ Step 5: Calculate Standard Deviation ($\sigma$) $\sigma = \sqrt{\sigma^2} = \sqrt{126.22} \approx 11.23$ Final Answer: Variance $\approx 126.22$ and Standard Deviation $\approx 11.23$.

Exam Tips for Statistics in CBSE Class 11

1. Formula Recall is Key: Many marks in this chapter depend on applying the correct formula. Create a 'formula sheet' for Mean Deviation (about mean/median, ungrouped/grouped data), Variance, and Standard Deviation. Practice writing them down without looking.
2. Accuracy in Calculations: Statistics problems often involve lengthy calculations. Use a calculator carefully and double-check your arithmetic, especially when dealing with decimals or large numbers. A small calculation error early on can cascade and lead to an incorrect final answer.
3. Understanding Absolute Values: For Mean Deviation, remember to take the absolute value of deviations (e.g., $|x_i - \bar{x}|$). A common mistake is to forget the absolute sign, leading to a sum of deviations that equals zero, which is incorrect for MD.
4. Squaring for Variance: When calculating variance, ensure you square the deviations $(x_i - \bar{x})$ before summing them up. This makes all values positive and prevents positive and negative deviations from cancelling each other out.
5. Square Root for Standard Deviation: Don't forget the final step of taking the square root to get the Standard Deviation from the Variance. It's a common oversight under exam pressure.
6. Distinguish Ungrouped vs. Grouped Data: Pay close attention to whether the given data is ungrouped (raw data) or grouped (frequency distribution). The formulas and steps differ significantly for each type.

Practice Questions with Solutions

• Q: Find the Mean Deviation about the Median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17. A: Step 1: Arrange the data in ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18. Step 2: Find the Median (M). There are 12 observations (even number). Median = (Value of 6th observation + Value of 7th observation) / 2 = (13 + 14) / 2 = 13.5. Step 3: Find absolute deviations from the median: $|x_i - M|$: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5. Step 4: Sum the absolute deviations: $\sum |x_i - M| = 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5 = 28. Step 5: Calculate Mean Deviation (MD): MD = $\frac{\sum |x_i - M|}{N} = \frac{28}{12} = \frac{7}{3} \approx 2.33$. Final answer: The Mean Deviation about the Median is approximately 2.33.
• Q: Calculate the variance for the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24. A: Step 1: Calculate the Mean ($\bar{x}$): $\bar{x} = \frac{6+8+10+12+14+16+18+20+22+24}{10} = \frac{150}{10} = 15$. Step 2: Calculate deviations $(x_i - \bar{x})$ and squared deviations $(x_i - \bar{x})^2$: $x_i - \bar{x}$: -9, -7, -5, -3, -1, 1, 3, 5, 7, 9 $(x_i - \bar{x})^2$: 81, 49, 25, 9, 1, 1, 9, 25, 49, 81. Step 3: Sum the squared deviations: $\sum (x_i - \bar{x})^2 = 81+49+25+9+1+1+9+25+49+81 = 330$. Step 4: Calculate Variance ($\sigma^2$): $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N} = \frac{330}{10} = 33$. Final answer: The Variance is 33.
• Q: Find the Standard Deviation for the data: 3, 5, 2, 7, 8, 4. A: Step 1: Calculate the Mean ($\bar{x}$): $\bar{x} = \frac{3+5+2+7+8+4}{6} = \frac{29}{6} \approx 4.83$. Step 2: Calculate deviations $(x_i - \bar{x})$ and squared deviations $(x_i - \bar{x})^2$: $x_i - \bar{x}$: -1.83, 0.17, -2.83, 2.17, 3.17, -0.83 $(x_i - \bar{x})^2$: 3.3489, 0.0289, 8.0089, 4.7089, 10.0489, 0.6889. Step 3: Sum the squared deviations: $\sum (x_i - \bar{x})^2 = 3.3489+0.0289+8.0089+4.7089+10.0489+0.6889 = 26.8334$. Step 4: Calculate Variance ($\sigma^2$): $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N} = \frac{26.8334}{6} \approx 4.4722$. Step 5: Calculate Standard Deviation ($\sigma$): $\sigma = \sqrt{\sigma^2} = \sqrt{4.4722} \approx 2.11$. Final answer: The Standard Deviation is approximately 2.11.
• Q: The mean and variance of 7 observations are 8 and 16, respectively. If each observation is multiplied by 3, find the new mean and new variance. A: Step 1: Original Mean ($\bar{x}$) = 8, Original Variance ($\sigma^2$) = 16, Number of observations (N) = 7. Step 2: Effect of multiplying by a constant on Mean: If each observation is multiplied by a constant 'k', the new mean will be 'k' times the original mean. New Mean = $3 \times \bar{x} = 3 \times 8 = 24$. Step 3: Effect of multiplying by a constant on Variance: If each observation is multiplied by a constant 'k', the new variance will be $k^2$ times the original variance. New Variance = $3^2 \times \sigma^2 = 9 \times 16 = 144$. Final answer: The new mean is 24 and the new variance is 144.

Frequently Asked Questions