NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2 Statistics - PDF Download
JEE Mains & AdvancedNCERT solutions for class 11 maths chapter 13 exercise 13.2 Statistics focuses on the second dispersion measure which is the standard deviation. Like the previous exercise, this concept is applicable to both grouped and ungrouped data. This section also provides an overview of the shortcut method for determining the variance and standard deviation. This method is also known as the step deviation method. It can be used when the size of a discrete distribution or the midpoints of different classes within a continuous distribution is large.
Class 11 maths chapter 13 exercise 13.2 NCERT solutions consists of 10 questions ranging from easy to intermediate levels. Students need to be able to use the formulas well to solve questions of this exercise. Ex 13.2 class 11 maths solutions are prepared by subject experts of eSaral. These solutions are explained in an easy step by step manner that helps you to excel in your exams. Ex 13.2 class 11 maths chapter 13 has been made available here in PDF format by eSaral’s experts to make your preparation easy and interesting. You can download the PDF for free from the official website of eSaral.
Topics Covered in Exercise 13.2 Class 11 Mathematics Questions
Ex 13.2 class 11 maths ch 13 is based on variance and standard deviation, standard deviation of a discrete frequency distribution and continuous frequency distribution, and a shortcut method to find variance and standard deviation. All these topics and concepts are described in a simple and precise way so that you can get the best out of it to solve questions of exercise 13.2 class 11 maths.
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Limitations of mean deviation |
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Variance and Standard Deviation |
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Limitations of mean deviation
In a series with a very high degree of variation, the median isn’t a representative central tendency, so the mean deviation about median calculated for this series can not be used as a full guide.
The deviation from the mean is more than the sum of deviation from the median. Therefore, the mean deviation about the mean is not very precise. Therefore, in most cases, the mean deviation may not give satisfactory results. Also, the mean deviation is calculated as an absolute value of the deviations. Therefore, it cannot be further analyzed in an algebraic way. This means that we need another measure of dispersion. Standard deviation is one such measure of dispersion.
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Variance and Standard Deviation
In statistics, the relationship between Variance and Standard Deviation is related because the standard deviation of a given data set is the square root of the variance.
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Variance - Variance is a measure of how widely a set of data is distributed. If all data values are the same, then it means that the variance is equal to zero. Any non-zero variance is considered positive. A small variance indicates that data points are near the mean and adjacent to each other. On the other hand, a high deviation means that data points are very distantly from each other and from the mean. In simple terms, the variance is the average squared distance from each point to the mean.
Variance formula for ungrouped data
$\sigma^2=\frac{1}{n} \sum\left(x_i-\bar{x}\right)^2$
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Standard Deviation - When calculating the variance, it is observed that the units for individual observations, xi, and the unit for their mean $\bar{x}$, are distinct from the unit for variance, as variance is the sum of the squares of (xi-$\bar{x}$). Therefore, the appropriate measure of dispersion about the mean of a set of observations is the positive square root of the variance, commonly referred to as the standard deviation. Therefore, the standard deviation, usually denoted by ‘σ’.
Standard deviation formula for ungrouped data
$\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2}$
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Standard deviation of a discrete frequency distribution
You can solve questions related to standard deviation of a discrete frequency distribution by the formula given-
$\sigma=\sqrt{\frac{1}{N} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2}$
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Standard deviation of a continuous frequency distribution - A discrete frequency distribution can be used to represent a given continuous frequency distribution by substituting each class for its midpoint. The standard deviation is then calculated using the same method as for a discrete frequency distribution. If n classes have a frequency distribution, each class is defined by its xi midpoint with a frequency of fi.
The standard deviation will be obtained by the formula-
$\sigma=\frac{1}{N} \sqrt{N \sum_{i=1}^n f_i x_i^2-\left(\sum_{i=1}^n f_i x_i\right)^2}$
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Shortcut method to find variance and standard deviation
Sometimes the xi values in the discrete distribution or the midpoint xi values of different classes in the continuous distribution are large, making the mean and variance calculation laborious and time-consuming. Step-deviation is a method of simplifying the process.
Shortcut method to find variance and standard deviation
$\sigma^2=\frac{h^2}{N^2}\left[n \sum f_i y_i^2-\left(\sum f_i y_i\right)^2\right]$, $\sigma=\frac{h}{N} \sqrt{N \sum f_i y_i^2-\left(\sum f_i y_i\right)^2}$
Where $y_i=\frac{x_i-A}{h}$
Tips for Solving Exercise 13.2 Class 11 Chapter 13 Statistics
NCERT solutions class 11 maths chapter 13 ex 13.2 Statistics involves the utilization of a variety of formulas. In order to gain an understanding of these formulas and how to apply them to various questions, students should follow the tips and techniques provided by academic teams of maths at eSaral.
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NCERT solutions class 11 maths chapter 13 ex 13.2 contains questions that are structured in different ways and require different techniques to solve the same problem. By solving these sums, students gain an understanding of the types of problems that may be encountered in an exam and can therefore plan their preparation accordingly.
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Students must learn and remember the formulas presented in this exercise to solve questions of ex 13.2 class 11 maths.
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To solve all the questions of exercise 13.2, you must go through each topic and concept related to variance and standard deviation.
Importance of Solving Ex 13.2 Class 11 Maths Chapter 13 Statistics
There are a lot of benefits of solving ex 13.2 questions of class 11 maths chapter 13 statistics. Here, our subject experts of eSaral have combined some of the benefits. You can check them below.
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Ex 13.2 class 11 maths chapter 13 NCERT solutions are prepared by expert faculty of eSaral which are explained in simple and precise language.
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NCERT solutions class 11 maths chapter 13 ex 13.2 has some significant concepts associated with variance and standard deviation which are solved in comprehensive and step by step manner.
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These solutions also include examples and essential questions which have been solved in an easy step by step way so that you can solve any question that comes in the exam.
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NCERT solutions are provided in PDF format that can be downloaded anytime, anywhere from eSaral website and practice each question with their answers.
Frequently Asked Questions
Question 1. What is standard deviation, discussed in NCERT solutions class 11 maths chapter 13 ex 13.2?
Answer 1. The standard deviation is a measure of the amount of variation or deviation that is present in a set of values. The range of values is determined by the standard deviation. In order to gain a better comprehension of the term, students can download PDF solutions offered by eSaral, which are offered in exercise-based formats to suit their individual requirements.
Question 2. What topics are discussed in NCERT solutions class 11 maths chapter 13 ex 13.2?
Answer 2. NCERT solutions class 11 maths chapter 13 ex 13.2 has some important topics such as variance and standard deviation, standard deviation, standard deviation of a discrete frequency distribution, standard deviation of a continuous frequency distribution, shortcut method to find variance and standard deviation and limitations of mean deviation which are explained by our expert teachers of eSaral. These solutions are also provided in PDF format which you can download on eSaral.