Vedantu’s RD Sharma Free PDF For Class 11 Solutions Available
FAQs on RD Sharma Class 11 Solutions Chapter 32 - Statistics (Ex 32.4) Exercise 32.4 - Free PDF
1. Where can I find accurate, step-by-step solutions for RD Sharma Class 11 Maths Exercise 32.4?
Vedantu provides detailed and expert-verified solutions for every problem in RD Sharma Class 11 Maths Chapter 32, Exercise 32.4. These solutions are crafted to explain the correct method for calculating variance and standard deviation, following the latest 2025-26 CBSE curriculum guidelines for clear understanding and exam preparation.
2. What is the correct method to calculate the standard deviation for a continuous frequency distribution in Exercise 32.4?
To calculate the standard deviation (σ) for a continuous frequency distribution, you should follow these steps precisely:
First, determine the mid-point (xᵢ) for each class interval.
Calculate the mean (μ) of the entire distribution using the formula μ = (Σfᵢxᵢ) / N.
Find the deviation of each mid-point from the mean (dᵢ = xᵢ - μ).
Square each of these deviations to get dᵢ².
Multiply each squared deviation by its corresponding frequency (fᵢdᵢ²).
The variance (σ²) is the mean of these values: σ² = (1/N) * Σfᵢdᵢ².
Finally, the standard deviation is the square root of the variance (σ = √σ²).
3. Which key formulas are essential for solving the problems in RD Sharma Class 11 Chapter 32, Exercise 32.4?
Exercise 32.4 primarily deals with variance and standard deviation for grouped data. The most crucial formulas are:
Variance (σ²): σ² = (1/N) * Σfᵢ(xᵢ - μ)²
Standard Deviation (σ): σ = √[(1/N) * Σfᵢ(xᵢ - μ)²]
Here, N is the total frequency (Σfᵢ), xᵢ represents the class mid-point, and μ is the distribution's mean. Understanding the shortcut or step-deviation method variations of these formulas is also vital.
4. Why is the step-deviation method often preferred for calculating variance in problems with large numbers?
The step-deviation method is preferred because it simplifies complex calculations. When dealing with large values for class mid-points (xᵢ) or frequencies, this method reduces the deviations into smaller, manageable integers by using an assumed mean and a common factor (h). This significantly lowers the chance of calculation errors and makes the process of finding variance and standard deviation much faster and more efficient.
5. What is a common mistake to avoid when calculating standard deviation for grouped data?
A frequent error students make is forgetting to multiply the squared deviation, (xᵢ - μ)², by its corresponding frequency (fᵢ). Forgetting this step is equivalent to treating the grouped data as ungrouped, leading to an incorrect result. Always ensure that the frequency of each class is factored into the sum of squared deviations, as the formula is Σfᵢ(xᵢ - μ)².
6. How does variance help in comparing the consistency of two different datasets?
Variance is a measure of the spread or dispersion of data points around the mean. When comparing two datasets, the one with the lower variance is considered more consistent because its data points are clustered more closely together. Conversely, a dataset with a higher variance indicates greater variability and less consistency. This concept is fundamental to analysing frequency distributions in Chapter 32.
7. In statistics, what is the practical meaning of a standard deviation of zero?
A standard deviation of zero signifies that there is no variability or spread in the data whatsoever. This is a special case where every single data point in the set is identical to the mean. For example, if every student in a class scores exactly 75 marks, the mean is 75 and the standard deviation is 0, indicating perfect consistency.






















