RD Sharma Class 11 Solutions Chapter 32 - Statistics (Ex 32.2) Exercise 32.2 - Free PDF
FAQs on RD Sharma Class 11 Solutions Chapter 32 - Statistics (Ex 32.2) Exercise 32.2
1. What is the primary concept a student must master to solve questions in RD Sharma Class 11 Solutions for Chapter 32, Exercise 32.2?
To successfully solve problems in this exercise, a student must master the calculation of mean deviation about the mean specifically for a discrete frequency distribution. This involves finding the weighted average of the absolute deviations of the observations from their mean, where the weights are the corresponding frequencies.
2. What is the correct step-by-step method to find the mean deviation about the mean for a discrete frequency distribution in Exercise 32.2?
To find the mean deviation about the mean for a discrete frequency distribution, follow these steps as per the methodology used in RD Sharma solutions:
Step 1: Calculate the mean (x̄) of the distribution using the formula x̄ = (Σfᵢxᵢ) / (Σfᵢ).
Step 2: Find the deviation of each observation (xᵢ) from the mean (x̄), which is (xᵢ - x̄).
Step 3: Take the absolute value of each deviation, i.e., |xᵢ - x̄|.
Step 4: Multiply each absolute deviation by its corresponding frequency (fᵢ) to get fᵢ|xᵢ - x̄|.
Step 5: Sum up all the values from the previous step to get Σfᵢ|xᵢ - x̄|.
Step 6: Divide this sum by the total number of observations (N = Σfᵢ) to get the final Mean Deviation.
3. How is solving for mean deviation in Exercise 32.2 different from the problems in Exercise 32.1?
The primary difference is the type of data. Exercise 32.1 deals with ungrouped data (individual observations), where each data point is considered individually. In contrast, Exercise 32.2 introduces discrete frequency distributions, where you work with observations (xᵢ) and their corresponding frequencies (fᵢ). This requires you to incorporate the frequency as a weight in your calculations for both the mean and the mean deviation, which is a key progression in complexity.
4. Why is taking the absolute value of deviations, |xᵢ - x̄|, a critical step when calculating mean deviation in this exercise?
Taking the absolute value is critical because the sum of simple deviations from the mean (Σ(xᵢ - x̄)) is always zero. This property would make it impossible to measure the actual spread or dispersion of data. By using absolute values, we ensure that all deviations are treated as positive distances from the mean, allowing us to calculate a meaningful average of these distances, which is what the mean deviation represents.
5. What is a common mistake to avoid while calculating the mean (x̄) for a discrete frequency distribution in Chapter 32?
A common mistake is forgetting to multiply each observation (xᵢ) by its corresponding frequency (fᵢ) before summing them up. Students sometimes calculate the mean by just summing the xᵢ values and dividing by the number of unique observations, instead of using the correct formula x̄ = Σfᵢxᵢ / Σfᵢ. The frequency of each data point must be factored in to find the true central tendency of the distribution.
6. Is it possible to calculate mean deviation about the median for the problems in Exercise 32.2? How would that change the approach?
Yes, it is possible. The approach would change significantly in the initial steps. Instead of first calculating the mean, you would need to find the median of the discrete frequency distribution. This involves finding the cumulative frequency and identifying the observation corresponding to the (N/2)th term, where N is the total frequency. After finding the median (M), all subsequent steps would use it as the central value, meaning you would calculate the sum of fᵢ|xᵢ - M| to find the mean deviation about the median.
7. How does mastering mean deviation in Exercise 32.2 help in understanding more advanced statistical concepts like variance?
Mastering mean deviation provides the foundational understanding of measuring data dispersion. The concept of summing the deviations from a central point is fundamental. Variance and standard deviation build directly on this idea but use the square of the deviations ((xᵢ - x̄)²) instead of the absolute value. This is done to give more weight to larger deviations and to create a measure with more convenient mathematical properties. Understanding mean deviation first makes the logic behind variance much easier to grasp.
