NCERT Solutions & Concepts: Analysis of Statistics Ex 15.1 Class 11

Why Do We Calculate Mean Deviation?

To understand dispersion, imagine two cricket players who have both scored an average of 50 runs over 10 matches. Player A's scores are consistently between 40 and 60, while Player B's scores range from 0 to 100. Although their average (mean) is identical, their reliability is very different. This variation is called dispersion.

Mean Deviation (M.D.) measures the average of the absolute differences between each data point and a central value (which can be the mean or the median). We take the absolute values (ignoring the negative signs) because the algebraic sum of deviations from the mean is always zero. By using absolute values, we get a true sense of the magnitude of spread from the center.

Essential Formulas for Exercise 15.1

Mean Deviation about Mean (Ungrouped)

M.D.(Mean) = [Σ |x_i - Mean|] / n, where Mean (x-bar) = (Σ x_i) / n and n is the number of observations.

Mean Deviation about Median (Ungrouped)

M.D.(Median) = [Σ |x_i - M|] / n, where M is the Median of the given data.

Mean Deviation about Mean (Grouped Data)

M.D.(Mean) = [Σ f_i * |x_i - Mean|] / N, where f_i is the frequency of observation x_i, and N = Σ f_i.

Mean Deviation about Median (Grouped Data)

M.D.(Median) = [Σ f_i * |x_i - M|] / N, where M is the Median and N = Σ f_i.

Step-by-Step Process to Find Mean Deviation

1. Step 1: Calculate the Central Tendency — Compute either the Mean (x-bar) or the Median (M) of the dataset depending on what the question demands.
2. Step 2: Find Absolute Deviations — Subtract the central value from each data point x_i. Take the absolute value: d_i = |x_i - Central Value|. Always treat negative values as positive.
3. Step 3: Multiply by Frequencies (if Grouped) — If working with grouped data, multiply each absolute deviation d_i by its corresponding frequency f_i to get f_i * d_i.
4. Step 4: Sum and Divide — Add all deviations (or product values) and divide the total sum by the total number of observations (n or N).

Common Mistakes & Board Exam Tips

• Arithmetic Errors: Statistics questions have long calculations. Double-check your basic additions and subtractions.
• Ignoring the Absolute Value: Students often forget to convert negative deviations to positive. Remember, | -5 | = 5.
• Wrong Median Selection: For ungrouped data, always arrange the data in ascending or descending order before calculating the median. Skipping this step leads to a wrong median.
• Grouping Midpoints: For continuous class intervals, remember to find the class midpoint (x_i = [Lower Limit + Upper Limit] / 2) first.

Practice Questions with Solutions

• Q: Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17. A: Step 1: Calculate the mean (x-bar). Mean = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17) / 8 = 80 / 8 = 10. Step 2: Calculate absolute deviations |x_i - Mean|: |4 - 10| = 6 |7 - 10| = 3 |8 - 10| = 2 |9 - 10| = 1 |10 - 10| = 0 |12 - 10| = 2 |13 - 10| = 3 |17 - 10| = 7 Step 3: Sum the absolute deviations: Σ |x_i - Mean| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24. Step 4: Divide by n (n = 8): M.D.(Mean) = 24 / 8 = 3. Final answer: The mean deviation about the mean is 3.
• Q: Find the mean deviation about the median for the data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. A: Step 1: Arrange data in ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21. Number of terms (n) = 11 (odd). Median (M) = ((11 + 1) / 2)th term = 6th term = 9. Step 2: Find absolute deviations |x_i - M|: |3-9|=6, |3-9|=6, |4-9|=5, |5-9|=4, |7-9|=2, |9-9|=0, |10-9|=1, |12-9|=3, |18-9|=9, |19-9|=10, |21-9|=12. Step 3: Sum the absolute deviations: Σ |x_i - M| = 6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12 = 58. Step 4: Compute M.D.(Median): M.D.(Median) = 58 / 11 ≈ 5.27. Final answer: The mean deviation about the median is approximately 5.27.
• Q: Find the mean deviation about the mean for the following discrete frequency distribution: x_i: 5, 10, 15, 20, 25 f_i: 7, 4, 6, 3, 5 A: Step 1: Find N = Σ f_i and Σ (f_i x_i). N = 7 + 4 + 6 + 3 + 5 = 25. Σ (f_i x_i) = (57) + (104) + (156) + (203) + (255) = 35 + 40 + 90 + 60 + 125 = 350. Mean (x-bar) = 350 / 25 = 14. Step 2: Find absolute deviations |x_i - Mean|: |5 - 14| = 9 |10 - 14| = 4 |15 - 14| = 1 |20 - 14| = 6 |25 - 14| = 11 Step 3: Calculate product f_i |x_i - Mean|: 7 9 = 63 4 4 = 16 6 1 = 6 3 6 = 18 5 * 11 = 55 Sum = 63 + 16 + 6 + 18 + 55 = 158. Step 4: M.D.(Mean) = 158 / 25 = 6.32. Final answer: The mean deviation about the mean is 6.32.
• Q: Why is Mean Deviation considered less mathematically stable compared to Standard Deviation? A: Step 1: Analyze the mathematical operations of both methods. Mean Deviation uses absolute values (modulus) to ignore signs. Step 2: Notice that absolute value functions are not differentiable at zero, which limits their use in algebraic treatments and advanced calculus. Step 3: Standard Deviation avoids this by squaring the differences instead of taking absolute values, making it highly suitable for further algebraic properties. Final answer: Mean deviation uses absolute values which are non-differentiable and mathematically rigid, making standard deviation the preferred measure in advanced mathematical statistics.

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