# NCERT Solutions for Class 11 Mathematics Chapter 15 Statistics

If you’re a Class 11 student studying Mathematics from the NCERT Textbook, you would have  probably come across Chapter 15 Statistics. Statistics is concerned with the gathering of information for specific purposes. Decisions can be made through analysing and interpreting facts. This chapter discusses measures of dispersion, range, mean deviation, variance, and standard deviation, as well as frequency distribution analysis.

You’ve already learned how to express data graphically and tabularly in previous classes. In Chapter 15 Class 11 Mathematics, you will learn about data representation, as well as specific prominent elements and qualities of the data. This is a chapter with a medium weightage for JEE Main.

# Class 11 Mathematics Chapter 15 – Topics

In the financial and economic industries, statistics is one of the most extensively used disciplines for planning and forecasting. They assist in the development of mathematical models and the analysis of time-series data. Therefore, students should extensively practise with the NCERT Solutions Class 12 Mathematics Chapter 15 and revise these concepts. The  solutions include answers, illustrations, and explanations for the entire chapter 15 titled Statistics, which is taught in Class 11.

Let’s go over the sections in this chapter before diving into the NCERT Solutions for Class 11 Mathematics Chapter 15:

 Exercise Topics 15.1 Introduction 15.2 Measures of Dispersion 15.3 Range 15.4 Mean Deviation 15.4.1 Mean Deviation for Ungrouped Data 15.4.2 Mean Deviation for Grouped Data 15.4.3 Limitations of Mean Deviation 15.5 Variance and Standard Deviation 15.5.1 Standard Deviation 15.5.2 Standard Deviation of a Discrete Frequency Distribution 15.5.3 Standard Deviation of a Continuous Frequency Distribution 15.5.4 Shortcut method to find Variance and Standard Deviation 15.6 Analysis of Frequency Distributions 15.6.1 Comparison of Two Frequency Distributions with same Mean

15.1 Introduction

With examples, this section discusses the concepts of central tendency, mean, and median [during even and odd number of observations]. It introduces the notion of dispersion measurement. Measures of central tendency are values that cluster around the middle or centre of the distribution. It includes the mean, the median, and the mode. In a class, mean can be used to calculate the average of the students’ grades.

15.2 Measures of Dispersion

This section discusses measurements of dispersion, including range, quartile deviation, mean deviation, and standard deviation.

Measures of dispersion explain the link with measures of central tendency. The spread of data, for example, indicates how well the data is represented by the mean. If the spread is high, the mean is not indicative of the data.

15.3 Range

This section explains the range, its calculation, and provides an example. The range represents the variety of scores by utilising the set’s highest and minimum values.

15.4 Mean Deviation

This section defines mean deviation and its formula. For example, biologists can utilise the notion of mean deviation to compare different animal weights and determine the weight that  is considered healthy.

15.4.1 Mean Deviation – Ungrouped  Data

This section describes how to calculate the mean deviation for ungrouped data. Students will have to calculate the mean, deviations from the mean, and absolute deviations, then enter the numbers into the mean deviation formula to get the result.

15.4.2 Mean Deviation – Grouped Data

This section explains how to calculate mean deviation for continuous and discrete frequency distributions using solved examples.

15.4.3 Mean deviation limitations

This section will cover the limitations of using mean deviation. Students will learn that:

• If there is a lot of variation in a series, the median will not be a good representation of the data. As a result, the mean deviation derived around such a median cannot be trusted.
• The mean deviation from the mean is not very specific if the sum of deviations from the mean is larger than the sum of deviations from the median.
• Further algebraic handling of the obtained absolute mean deviation is not possible. It cannot be used as an accurate measure of dispersion.

15.5 Standard Deviation and Variance

15.5.1 Standard Deviation

This section defines variance and standard deviation, as well as provides formulae  and solved examples for discrete and continuous frequency distributions.

15.5.2 Standard Deviation of a Discrete Frequency Distribution

Students will learn that standard deviation can be determined by one of two methods based on the frequency distribution of the dataset. One of such methods is when the pieces of the data presented are discontinuous (discrete) in which there are no observation groups, and each element is assigned a specific frequency value.

15.5.3. Standard Deviation of a Continuous Frequency Distribution

The second method is when the data is provided, i.e., the elements or observations are displayed in a continuous or grouped format. It means that instead of providing a specific value to each frequency, a ‘class’ or ‘group’ of a specific element range is provided.

15.5.4 Quick Technique for Calculating Variance and Standard Deviation

With a few illustrations, this section discusses the simpler method of determining the standard deviation.

15.6 Frequency Distribution Analysis

This section explains how to compare the variability of two series with the same mean, coefficient of variation, and a few solved issues.

15.6.1 Comparison of Two Frequency Distributions with Same Mean

To compare the variability or dispersion of two series, the coefficient of variance for each series should be estimated. It is calculated by dividing the standard deviation by the mean and finding its percentage.

### NCERT Solutions for Class 11 Mathematics Chapter 15 – Statistics

You must be seeking answers to the questions after you have completed Chapter 15 Mathematics Class 11. The solutions provide alternative solutions and clarifications, making students feel more confident when appearing in the exam. In case, students aren’t able to derive the answer or are unsure if their answer/approach is correct, solutions by Extramarks will prove to be of great help. The subject matter experts from Extramarks have designed the Class 11 Mathematics NCERT Solutions Chapter 15 based on the most recent CBSE Syllabus 2022-23, keeping in mind the question paper setting and the distribution of marks throughout the chapters. You can access and download these solutions and practise them offline as well.

Access NCERT Solutions for Class 11 Mathematics Chapter 15 – Statistics

SOLUTIONS

### Class 11 Mathematics NCERT Solutions Chapter 15

As students move through their educational journey, the concepts provided in this chapter will serve as the foundation for the introduction of more complex topics. As a result, students should make it a point to be well-versed in the formulas and theories of Statistics, as these will undoubtedly make their way into future board exams. These NCERT Solutions offer a variety of practical examples to deliver an engaging lesson, guaranteeing that students gain a solid statistical foundation and also download the exercise-by-exercise solutions given in the links below.

SOLUTIONS

In the following chapter, we go over all of the exercises:

How is Statistics Class 11 Mathematics NCERT Solutions Advantageous to Students?

If you ask any top student what their secret to success is, they will undoubtedly answer that they extensively read the NCERT textbooks. All of the fundamentals are covered in these textbooks. After you have finished the theories, it is critical that you solve the NCERT questions thoroughly. This is essential for both – board exams and competitive tests such as JEE Main and JEE Advanced.

The NCERT Solutions for Class 11 Mathematics Chapter 15 have exact explanations in easy language to assist students in performing well in their first term exams. Before moving on to more difficult topics, basic concepts are thoroughly explained so that students’ doubts or confusion are cleared.

The step-by-step manner of problem-solving offers students a thorough understanding of the mark weightage as per the new CBSE Syllabus 2022-23. Students will be able to identify their areas of weakness and try to improve them in order to improve their academic performance.

You now have access to all of the Class 11 Mathematics NCERT Solutions Chapter 15 on Statistics. Use these solutions to your advantage and master the chapter.

Q.1 Find the mean deviation about the mean for the data in Exercises 1 and 2.
Q.1: 4, 7, 8, 9, 10, 12, 13, 17
Q.2: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans

1.

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Mean}=\frac{4+7+8+9+10+12+13+17}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\\ \mathrm{Mean}\text{deviation}=\left\{\frac{\begin{array}{l}|4-10|+|7-10|+|8-10|+|9-10|+|10-10|\\ +|12-10|+|13-10|+|17-10|\end{array}}{8}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{6+3+2+1+0+2+3+7}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{24}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\end{array}$

Thus, the mean deviation is 3.

2.

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Mean}=\frac{38+70+48+40+42+55+63+46+54+44}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{500}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\\ \mathrm{Mean}\text{deviation}=\left\{\frac{\begin{array}{l}|38-50|+|70-50|+|48-50|+|40-50|+\\ |42-50|+|55-50|+|63-50|+|46-50|+\\ |54-50|+|44-50|\end{array}}{10}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{12+20+2+10+8+5+13+4+4+6}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{84}{10}\text{\hspace{0.17em}}=8.4\\ \text{Thus},\text{the mean deviation is 8.4}.\end{array}$

Q.2 Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Ans

$\begin{array}{l}3.\mathrm{A}\mathrm{scending}\text{order of data:}\\ 10,11,11,12,13,13,14,16,16,17,17,18\\ \mathrm{Here},\text{n}=12\left(\mathrm{even}\right)\\ \mathrm{Median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\end{array}$ $\begin{array}{ll}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{12}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{12}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{6}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}+{7}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}}{2}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{13+14}{2}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{27}{2}& \phantom{\rule{0ex}{0ex}}\\ \mathrm{Median}=13.5& \phantom{\rule{0ex}{0ex}}\\ \mathrm{Mean}\text{deviation about median}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\frac{\begin{array}{l}|10-13.5|+|11-13.5|+|11-13.5|+|12-13.5|+\\ |13-13.5|+|13-13.5|+|14-13.5|+|16-13.5|+\\ |16-13.5|+|17-13.5|+|17-13.5|+|18-13.5|\end{array}}{12}\right\}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\begin{array}{l}3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5\\ +3.5+3.5+4.5\end{array}\right)}{12}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{28}{12}& \phantom{\rule{0ex}{0ex}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2.33\left(\mathrm{Approx}.\right)& \end{array}$ $\begin{array}{l}4.\text{}\mathrm{Ascending}\text{order of data:}\\ 36,42,45,46,46,49,51,53,60,72\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Here},\text{n}=10\left(\mathrm{even}\right)\\ \mathrm{Median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{10}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{10}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{5}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}+{6}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{46+49}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{95}{2}\\ \mathrm{Median}=47.5\\ \mathrm{Mean}\text{deviation about median}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\frac{\begin{array}{l}|36-47.5|+|42-47.5|+|45-47.5|+|46-47.5|+\\ |46-47.5|+|49-47.5|+|51-47.5|+|53-47.5|+\\ |60-47.5|+|72-47.5|\end{array}}{10}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\begin{array}{l}11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5\\ +12.5+24.5\end{array}\right)}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{70}{10}\text{\hspace{0.17em}}=7\end{array}$

Q.3 5. Find the mean deviation about the mean for the data in Exercises 5 and 6.

 Xi 5 10 15 20 25 fi 7 4 6 3 5

6.

 Xi 10 30 50 70 90 fi 4 24 28 16 8

Ans

$\begin{array}{|ccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}& \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|& {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\\ 5& 7& 35& \left|5-14\right|=9& 63\\ 10& 4& 40& \left|10-14\right|=4& 16\\ 15& 6& 90& \left|15-14\right|=1& 6\\ 20& 3& 60& \left|20-14\right|=6& 18\\ 25& 5& 125& \left|25-14\right|=11& 55\\ \mathrm{Total}& 25& 350& & 158\\ \hline\end{array}$ $\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{350}{25}\\ \text{ }=14\\ \mathrm{Mean}\text{deviation about mean}\\ \text{ }=\frac{\sum {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{158}{25}=6.32\\ \mathrm{Thus},\text{the mean deviation about mean is 6.32.}\end{array}$ $\begin{array}{|ccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}& \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|& {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\\ 10& 4& 40& \left|10-50\right|=40& 160\\ 30& 24& 720& \left|30-50\right|=20& 480\\ 50& 28& 1400& \left|50-50\right|=0& 0\\ 70& 16& 1120& \left|70-50\right|=20& 320\\ 90& 8& 720& \left|90-50\right|=40& 320\\ \mathrm{Total}& 80& 4000& & 1280\\ \hline\end{array}$ $\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{4000}{80}\\ \text{ }=50\\ \mathrm{Mean}\text{deviation about mean}\\ \text{ }=\frac{\sum {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{1280}{80}=16\\ \mathrm{Thus},\text{the mean deviation about mean is 16.}\end{array}$

Q.4 Find the mean deviation about the median for the data in Exercises 7 and 8.

7.
 Xi 5 7 9 10 12 15 fi 8 6 2 2 2 6
8.
 Xi 15 21 27 30 35 fi 3 5 6 7 8

Ans

7.
$\begin{array}{|ccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& \mathrm{C}.\mathrm{f}.& |{\mathrm{x}}_{\mathrm{i}}-\mathrm{Median}|& {\mathrm{f}}_{\mathrm{i}}|{\mathrm{x}}_{\mathrm{i}}-\mathrm{Median}|\\ 5& 8& 8& |5-7|=2& 16\\ 7& 6& 14& |7-7|=0& 0\\ 9& 2& 16& |9-7|=2& 4\\ 10& 2& 18& |10-7|=3& 6\\ 12& 2& 20& |12-7|=5& 10\\ 15& 6& 26& |15-7|=8& 48\\ \mathrm{Total}& 26& & & 84\\ \hline\end{array}$

$\begin{array}{l}\text{n = 26}\left(\text{even}\right)\\ \text{Median = }\frac{{\left(\frac{\text{n}}{\text{2}}\right)}^{\text{th}}\text{\hspace{0.17em}term+}{\left(\frac{\text{n}}{\text{2}}\text{+1}\right)}^{\text{th}}\text{\hspace{0.17em}term}}{\text{2}}\\ \text{ = }\frac{{\left(\frac{\text{26}}{\text{2}}\right)}^{\text{th}}\text{\hspace{0.17em}term+}{\left(\frac{\text{26}}{\text{2}}\text{+1}\right)}^{\text{th}}\text{\hspace{0.17em}term}}{\text{2}}\\ \text{ = }\frac{{\text{13}}^{\text{th}}{\text{\hspace{0.17em}term+14}}^{\text{th}}\text{\hspace{0.17em}term}}{\text{2}}\\ \text{ = }\frac{\text{7+7}}{\text{2}}\\ \text{ = 7}\\ \text{Mean deviation about median}\\ \text{ = }\frac{\sum {\text{f}}_{\text{i}}\left|{\text{x}}_{\text{i}}\text{-Me}\right|}{\sum {\text{f}}_{\text{i}}}\\ \text{ = }\frac{\text{84}}{\text{26}}\\ \text{ = 3.23}\\ \text{Thus, Mean deviation about median is 3.23.}\end{array}$

8.

$\begin{array}{|ccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& \mathrm{C}.\mathrm{f}.& \left|{\mathrm{x}}_{\mathrm{i}}-\mathrm{Median}\right|& {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\mathrm{Median}\right|\\ 15& 3& 3& \left|15-30\right|=15& 45\\ 21& 5& 8& \left|21-30\right|=9& 45\\ 27& 6& 14& \left|27-30\right|=3& 18\\ 30& 7& 21& \left|30-30\right|=0& 0\\ 35& 8& 29& \left|35-30\right|=5& 40\\ \mathrm{Total}& 29& & & 148\\ \hline\end{array}$ $\begin{array}{l}\text{n = 29}\left(\text{odd}\right)\\ \text{Median= }{\left(\frac{\text{n+1}}{\text{2}}\right)}^{\text{th}}\text{\hspace{0.17em}term}\\ \text{ = }{\left(\frac{\text{29+1}}{\text{2}}\right)}^{\text{th}}\text{\hspace{0.17em}term}\\ {\text{ = 15}}^{\text{th}}\text{\hspace{0.17em}term}\\ \text{ = 30}\\ \text{Mean deviation about Median}\\ \text{ = }\frac{\sum {\text{f}}_{\text{i}}\left|{\text{x}}_{\text{i}}\text{-Me}\right|}{\sum {\text{f}}_{\text{i}}}\\ \text{ = }\frac{\text{148}}{\text{29}}\\ \text{ = 5.1}\\ \text{Thus, the Mean deviation about Median is 5.1.}\end{array}$

Q.5 Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

 Income per day 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 Number of persons 4 8 9 10 7 5 4 3
10.
 Height in cms 95-105 105-115 115-125 125-135 135-145 145-155 Number of boys 9 13 26 30 12 10

Ans

9.

$\begin{array}{|ccccccc|}\hline \mathrm{Income}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& \begin{array}{l}{\mathrm{u}}_{\mathrm{i}}\\ =\frac{\mathrm{x}-\mathrm{A}}{\mathrm{h}}\end{array}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|& {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\\ 0-100& 4& 50& -4& -16& \left|50-358\right|=308& 1232\\ 100-200& 8& 150& -3& -24& \left|150-358\right|=208& 1664\\ 200-300& 9& 250& -2& -18& \left|250-358\right|=108& 972\\ 300-400& 10& 350& -1& -10& \left|350-358\right|=8& 80\\ 400-500& 7& 450=\mathrm{A}& 0& 0& \left|450-358\right|=92& 644\\ 500-600& 5& 550& 1& 5& \left|550-358\right|=192& 960\\ 600-700& 4& 650& 2& 8& \left|650-358\right|=292& 1168\\ 700-800& 3& 750& 3& 9& \left|750-358\right|=392& 1176\\ \mathrm{Total}& 50& & -46& & & 7896\\ \hline\end{array}$ $\begin{array}{l}\mathrm{Mean}=\mathrm{A}+\mathrm{h}×\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}},{\text{where u}}_{\text{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{\mathrm{h}}\text{and h}=\text{100}\\ \text{ }=450+100×\frac{-46}{50}\\ \text{ }=450-2×46\\ \text{ }=450-92\\ \text{ }=358\\ \mathrm{Mean}\text{deviation about mean}\\ \text{ }=\frac{\sum {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{7896}{50}\\ \text{ }=157.92\end{array}$

10.

$\begin{array}{|ccccccc|}\hline \begin{array}{l}\mathrm{Class}-\\ \mathrm{interval}\end{array}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{\mathrm{h}}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|& {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\\ 95-105& 9& 100& -3& -27& \left|100-125.3\right|=25.3& 227.7\\ 105-115& 13& 110& -2& -26& \left|110-125.3\right|=15.3& 196.9\\ 115-125& 26& 120& -1& -26& \left|120-125.3\right|=5.3& 137.8\\ 125-135& 30& 130=\mathrm{A}& 0& 0& \left|130-125.3\right|=4.7& 141\\ 135-145& 12& 140& 1& 12& \left|140-125.3\right|=14.7& 176.4\\ 145-155& 10& 150& 2& 20& \left|150-125.3\right|=24.7& 247\\ \mathrm{Total}& 100& & & -47& & 1128.8\\ \hline\end{array}$ $\begin{array}{l}\mathrm{Mean}=\mathrm{A}+\mathrm{h}×\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}},{\text{where u}}_{\text{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{\mathrm{h}}\text{and h}=\text{10}\\ \text{ }=130+10×\frac{-47}{100}\\ \text{ }=130-4.7\\ \text{ }=125.3\\ \mathrm{Mean}\text{deviation about Mean}\\ \text{ }=\frac{\sum {\mathrm{f}}_{\mathrm{i}}\left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{ }=\frac{1128.8}{100}\\ \text{ }=11.288\\ \text{ }=11.29\left(\mathrm{Approx}\right)\\ \mathrm{Thus},\text{}\mathrm{Mean}\text{deviation about Mean is 11.29.}\end{array}$

Q.6 Find the mean deviation about median for the following data :

 Marks 0-10 Oct-20 20-30 30-40 40-50 50-60 Number of Girls 6 8 14 16 4 2

Ans

$\begin{array}{l}\begin{array}{|cccccc|}\hline \mathrm{Marks}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& \mathrm{c}.\mathrm{f}.& |{\mathrm{x}}_{\mathrm{i}}-\mathrm{Me}|& {\mathrm{f}}_{\mathrm{i}}|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}|\\ 0-10& 6& 5& 6& |5-27.85|=22.85& 137.1\\ 10-20& 8& 15& 14& |15-27.85|=12.85& 102.8\\ 20-30& 14& 25& 28& |25-27.85|=2.85& 39.9\\ 30-40& 16& 35& 44& |35-27.85|=7.15& 114.4\\ 40-50& 4& 45& 48& |45-27.85|=17.15& 68.6\\ 50-60& 2& 55& 50& |55-27.85|=27.15& 54.3\\ \mathrm{Total}& 50& & & & 517.1\\ \hline\end{array}\\ \mathrm{N}=50\left(\mathrm{even}\right)\\ \frac{\mathrm{N}}{2}=\frac{50}{2}=25\\ \mathrm{So},\text{the median class is 20}-\text{30.}\\ \mathrm{Median}=\mathrm{l}+\frac{\left(\frac{\mathrm{N}}{2}\right)-\mathrm{C}}{\mathrm{f}}×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20+\frac{25-14}{14}×10\left[\because \mathrm{C}=14,\mathrm{f}=14\text{​\hspace{0.17em}\hspace{0.17em}and h=10}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20+7.85=27.85\\ \mathrm{Mean}\text{deviation about Meadian}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{517.1}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}}=10.34\\ \mathrm{Thus},\text{}\mathrm{Mean}\text{deviation about Meadian is 10.34.}\end{array}$

Q.7 Calculate the mean deviation about median age for the age distribution of 100 persons given below:

 Age 16-20 21-25 36-30 31-35 36-40 41-45 46-50 51-55 Number 5 6 12 14 26 12 16 9

Ans

The given table can be arranged as given below:

$\begin{array}{|cccccc|}\hline \mathrm{Marks}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& \mathrm{c}.\mathrm{f}.& |{\mathrm{x}}_{\mathrm{i}}-\mathrm{Me}|& {\mathrm{f}}_{\mathrm{i}}|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}|\\ 15.5-20.5& 5& 18& 5& |18-38|=20& 100\\ 20.5-25.5& 6& 23& 11& |23-38|=15& 90\\ 25.5-30.5& 12& 28& 23& |28-38|=10& 120\\ 30.5-35.5& 14& 33& 37& |33-38|=5& 70\\ 35.5-40.5& 26& 38& 63& |38-38|=0& 0\\ 40.5-45.5& 12& 43& 75& |43-38|=5& 60\\ 45.5-50.5& 16& 48& 91& |48-38|=10& 160\\ 50.5-55.5& 9& 53& 100& |53-38|=15& 135\\ \mathrm{Total}& 100& & & & 735\\ \hline\end{array}$ $\begin{array}{l}\mathrm{N}=100\left(\mathrm{even}\right)\\ \frac{\mathrm{N}}{2}=\frac{100}{2}=50\\ \mathrm{So},\text{the median class is}35.5-40.5.\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\mathrm{Median}=\mathrm{l}+\frac{\left(\frac{\mathrm{N}}{2}\right)-\mathrm{C}}{\mathrm{f}}×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35.5+\frac{50-37}{26}×5\left[\because \mathrm{C}=37,\mathrm{f}=26\text{​\hspace{0.17em}\hspace{0.17em}and h}=\text{5}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35.5+2.5=38\\ \mathrm{Mean}\text{deviation about Meadian}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sum {\mathrm{f}}_{\mathrm{i}}|{\mathrm{x}}_{\mathrm{i}}-\mathrm{Me}|}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{735}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}}=7.35\\ \mathrm{Thus},\text{}\mathrm{Mean}\text{deviation about Meadian is 7.35.}\end{array}$

Q.8 Find the mean and variance for each of the data in Exercies 1 to 5.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
Q2. First n natural numbers.
Q3. First 10 multiples of 3.
Q4.

 Xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3
Q5.
 Xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

Ans

1.

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{6+7+10+12+13+4+8+12}{8}\\ =\frac{72}{8}\\ =9\end{array}$ $\begin{array}{l}\begin{array}{|ccc|}\hline {\mathrm{x}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 6& 6-9=-3& 9\\ 7& 7-9=-2& 4\\ 10& 10-9=1& 1\\ 12& 12-9=3& 9\\ 13& 13-9=4& 16\\ 4& 4-9=-5& 25\\ 8& 8-9=-1& 1\\ 12& 12-9=3& 9\\ & \mathrm{Total}& 74\\ \hline\end{array}\\ \text{\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{N}}\\ =\frac{74}{8}\\ =9.25\end{array}$

2.

$\begin{array}{l}\mathrm{Mean}\text{of n natural numbers}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}\\ =\frac{\left(\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right)}{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{{\mathrm{x}}_{\mathrm{i}}}^{2}-{\left(\overline{\mathrm{x}}\right)}^{2}}{\mathrm{n}}\end{array}$ $\begin{array}{l}=\frac{\left({1}^{2}+{2}^{2}+{3}^{2}+...+{\mathrm{n}}^{2}\right)}{\mathrm{n}}-{\left(\frac{\mathrm{n}+1}{2}\right)}^{2}\\ =\frac{1}{\mathrm{n}}×\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-\left(\frac{{\mathrm{n}}^{2}+2\mathrm{n}+1}{4}\right)\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{2{\mathrm{n}}^{2}+3\mathrm{n}+1}{6}-\left(\frac{{\mathrm{n}}^{2}+2\mathrm{n}+1}{4}\right)\\ =\frac{4{\mathrm{n}}^{2}+6\mathrm{n}+2-3{\mathrm{n}}^{2}-6\mathrm{n}-3}{12}\\ =\frac{{\mathrm{n}}^{2}-1}{12}\end{array}$

3.

$\begin{array}{l}\mathrm{Mean}\text{of 10 multiples of 3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\begin{array}{l}3+6+9+12+15+18+21+24\\ +27+30\end{array}\right)}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{165}{10}=16.5\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{{\mathrm{x}}_{\mathrm{i}}}^{2}-{\left(\overline{\mathrm{x}}\right)}^{2}}{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\begin{array}{l}{3}^{2}+{6}^{2}+{9}^{2}+{12}^{2}+{15}^{2}+{18}^{2}\\ +{21}^{2}+{24}^{2}+{27}^{2}+{30}^{2}\end{array}\right)}{10}-{\left(16.5\right)}^{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3465}{10}-272.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=346.5-272.25\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)\text{\hspace{0.17em}}=74.25\end{array}$

4.

$\begin{array}{l}\begin{array}{|cccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}& {\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 6& 2& 12& 6-19=-13& 169& 338\\ 10& 4& 40& 10-19=-9& 81& 324\\ 14& 7& 98& 14-19=-5& 25& 175\\ 18& 12& 216& 18-19=-1& 1& 12\\ 24& 8& 192& 24-19=5& 25& 200\\ 28& 4& 112& 28-19=9& 81& 324\\ 30& 3& 90& 30-19=11& 121& 363\\ \mathrm{Total}& 40& 760& & & 1736\\ \hline\end{array}\\ \mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}\\ =\frac{760}{40}\\ =19\end{array}$ $\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{n}}$ $\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{1736}{40}\\ =43.4\end{array}$

5.

$\begin{array}{l}\begin{array}{|cccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}& {\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 92& 3& 276& 92-100=-8& 64& 192\\ 93& 2& 186& 93-100=-7& 49& 98\\ 97& 3& 291& 97-100=-3& 9& 27\\ 98& 2& 196& 98-100=-2& 4& 8\\ 102& 6& 612& 102-100=2& 4& 24\\ 104& 3& 312& 104-100=4& 16& 48\\ 109& 3& 327& 109-100=9& 81& 243\\ \mathrm{Total}& 22& 2200& & & 640\\ \hline\end{array}\\ \mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}\\ =\frac{2200}{22}\\ =100\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{n}}\end{array}$ $\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{640}{22}\\ =29.09\end{array}$

Q.9 Find the mean and standard deviation using short-cut method.

 Xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5

Ans

$\begin{array}{|ccccccc|}\hline {\mathrm{x}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{d}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{1}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}& {\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 60& 2& -\text{\hspace{0.17em}}4& -\text{\hspace{0.17em}}8& -4& 16& 32\\ 61& 1& -\text{\hspace{0.17em}}3& -3& -3& 9& 9\\ 62& 12& -\text{\hspace{0.17em}}2& -\text{\hspace{0.17em}}24& -2& 4& 48\\ 63& 29& -\text{\hspace{0.17em}}1& -29& -1& 1& 29\\ 64=\mathrm{A}& 25& 0& 0& 0& 0& 0\\ 65& 12& 1& 12& 1& 1& 12\\ 66& 10& 2& 20& 2& 4& 40\\ 67& 4& 3& 12& 3& 9& 36\\ 68& 5& 4& 20& 4& 16& 80\\ \mathrm{Total}& 100& & 0& & & 286\\ \hline\end{array}$ $\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\mathrm{n}}\\ \text{ }=64+\frac{0}{100}=64\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{n}}\\ \text{ }=\frac{286}{100}\\ \text{ }=2.86\\ \text{Standard deviation}\\ \text{ }=\sqrt{2.86}\\ \text{ }=1.69\left(\mathrm{Approx}.\right)\end{array}$

Q.10 Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Q7.

 Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2
Q8.
 Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6

Ans

7.

$\begin{array}{|ccccccc|}\hline \mathrm{Classes}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{30}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 0-30& 2& 15& -3& 9& -6& 18\\ 30-60& 3& 45& -2& 4& -6& 12\\ 60-90& 5& 75& -1& 1& -5& 5\\ 90-120& 10& 105=\mathrm{A}& 0& 0& 0& 0\\ 120-150& 3& 135& 1& 1& 3& 3\\ 150-180& 5& 165& 2& 4& 10& 20\\ 180-210& 2& 195& 3& 9& 6& 18\\ \mathrm{Total}& 30& & & & 2& 76\\ \hline\end{array}$ $\begin{array}{l}\text{}\text{}Mean=A+\frac{\sum _{i=1}^{n}{f}_{i}{u}_{i}}{n}×h\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=105+\frac{2}{30}×30\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=107\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(30\right)}^{2}}{{\left(30\right)}^{2}}\left[30×76-{\left(2\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=2280-4=2276\end{array}$

8.

$\begin{array}{l}\begin{array}{|ccccccc|}\hline \mathrm{Classes}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{10}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 0-10& 5& 5& -2& 4& -10& 20\\ 10-20& 8& 15& -1& 1& -8& 8\\ 20-30& 15& 25=\mathrm{A}& 0& 0& 0& 0\\ 30-40& 16& 35& 1& 1& 16& 16\\ 40-50& 6& 45& 2& 4& 12& 24\\ \mathrm{Total}& 50& & & & 10& 68\\ \hline\end{array}\\ \mathrm{Mean}=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25+\frac{10}{50}×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(10\right)}^{2}}{{\left(50\right)}^{2}}\left[50×68-{\left(10\right)}^{2}\right]\\ =\frac{1}{25}\left[3400-100\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=132\end{array}$

Q.11 Find the mean, variance and standard deviation using short-cut method

 Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of children 3 4 7 7 15 9 6 6 3

Ans

$\begin{array}{|ccccccc|}\hline Classes& {f}_{i}& {x}_{i}& {u}_{i}=\frac{{x}_{i}-A}{5}& {u}_{i}{}^{2}& {f}_{i}{u}_{i}& {f}_{i}{u}_{i}{}^{2}\\ 70-75& 3& 72.5& -4& 16& -12& 48\\ 75-80& 4& 77.5& -3& 9& -12& 36\\ 80-85& 7& 82.5& -2& 4& -14& 28\\ 85-90& 7& 87.5& -1& 1& -7& 7\\ 90-95& 15& 92.5=A& 0& 0& 0& 0\\ 95-100& 9& 97.5& 1& 1& 9& 9\\ 100-105& 6& 102.5& 2& 4& 12& 24\\ 105-110& 6& 107.5& 3& 9& 18& 54\\ 110-115& 3& 112.5& 4& 16& 12& 78\\ Total& 60& & & & 6& 254\\ \hline\end{array}$ $\mathrm{Mean}=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{}\end{array}$ $\begin{array}{l}\text{}=92.5+\frac{6}{60}×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=92.5+0.5\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=93\\ \text{\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \text{\hspace{0.17em} \hspace{0.17em}}=\frac{{\left(5\right)}^{2}}{{\left(60\right)}^{2}}\left[60×254-{\left(6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{3600}\left[15240-36\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{3600}×15204\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=105.52\\ \text{Standard deviation}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\sqrt{105.52}\\ \text{\hspace{0.17em}\hspace{0.17em}}=10.27\end{array}$

Q.12 The diameters of circles (in mm) drawn in a design are given below:

 Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.

Ans

$\begin{array}{l}\begin{array}{|ccccccc|}\hline \begin{array}{c}\text{Classe-interva}\end{array}& \begin{array}{c}{\text{f}}_{\text{i}}\end{array}& \begin{array}{c}{\text{x}}_{\text{i}}\end{array}& {\text{u}}_{\text{i}}\text{=}\frac{{\text{x}}_{\text{i}}\text{-A}}{\text{4}}& {{\text{u}}_{\text{i}}}^{\text{2}}& {\text{f}}_{\text{i}}\text{u}& {\text{f}}_{\text{i}}{{\text{u}}_{\text{i}}}^{\text{2}}\\ \text{32.5-36.5}& \text{15}& \text{34.5}& \text{-2}& \text{4}& \text{-30}& \text{60}\\ \text{36.5-40.5}& \text{17}& \text{38.5}& \text{-1}& \text{1}& \text{-17}& \text{-17}\\ \text{40.5-44.5}& \text{21}& \text{42.5=A}& \text{0}& \text{0}& \text{0}& \text{0}\\ \text{44.5-48.5}& \text{22}& \text{46.5}& \text{1}& \text{1}& \text{22}& \text{22}\\ \text{48.5-52.5}& \text{25}& \text{50.5}& \text{2}& \text{4}& \text{50}& \text{100}\\ \text{Total}& \text{100}& & & & \text{25}& \text{199}\\ \hline\end{array}\\ \text{ Variance}\left({\text{σ}}^{\text{2}}\right)\text{ = }\frac{{\text{h}}^{\text{2}}}{{\text{n}}^{\text{2}}}\left[\text{n}\sum _{\text{i}}^{\text{n}}\left({\text{f}}_{\text{i}}{{\text{u}}_{\text{i}}}^{\text{2}}\right)\text{–}{\left(\sum _{\text{i}}^{\text{n}}{\text{f}}_{\text{i}}{\text{u}}_{\text{i}}\right)}^{\text{2}}\right]\\ \text{ = }\frac{{\left(\text{4}\right)}^{\text{2}}}{{\left(\text{100}\right)}^{\text{2}}}\left[\text{100×199-}{\left(\text{25}\right)}^{\text{2}}\right]\\ \text{ = }\frac{\text{1}}{\text{625}}\left[\text{19900-625}\right]\\ \text{ = }\frac{\text{1}}{\text{625}}\text{×19275}\\ \text{ = 30.84}\\ \text{Standard deviation}\\ \text{ = }\sqrt{\text{30.84}}\\ \text{ = 5.55}\\ \text{ Mean = A+}\frac{\sum _{\text{i=1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{u}}_{\text{i}}}{\text{n}}\text{×h}\\ \text{ = 42.5+}\frac{\text{25}}{\text{100}}\text{×4}\\ \text{ = 42.5+1}\\ \text{ = 43.5}\end{array}$

Q.13 From the data given below state which group is more variable, A or B?

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

Ans

For group A:

$\begin{array}{l}\begin{array}{|ccccccc|}\hline \mathrm{Marks}\left(\mathrm{A}\right)& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{10}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 10-20& 9& 15& -3& 9& -27& 81\\ 20-30& 17& 25& -2& 4& -34& 68\\ 30-40& 32& 35& -1& 1& -32& 32\\ 40-50& 33& 45=\mathrm{A}& 0& 0& 0& 0\\ 50-60& 40& 55& 1& 1& 40& 40\\ 60-70& 10& 65& 2& 4& 20& 40\\ 70-80& 9& 75& 3& 9& 27& 81\\ & 150& & & & -6& 342\\ \hline\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}\\ \text{\hspace{0.17em}}=45+\frac{-6}{150}×10\\ \text{\hspace{0.17em}}=45-0.4\text{\hspace{0.17em}}=44.6\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\end{array}$ $\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\left(10\right)}^{2}}{{\left(150\right)}^{2}}\left[150×342-{\left(-6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}\left(51300-36\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}×51264\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=227.84\\ \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{227.84}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15.09\end{array}$ $\begin{array}{|ccccccc|}\hline \mathrm{Marks}\left(\mathrm{A}\right)& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{10}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 10-20& 10& 15& -3& 9& -30& 90\\ 20-30& 20& 25& -2& 4& -\text{\hspace{0.17em}}40& 80\\ 30-40& 30& 35& -1& 1& -30& 30\\ 40-50& 25& 45=\mathrm{A}& 0& 0& 0& 0\\ 50-60& 43& 55& 1& 1& 43& 43\\ 60-70& 15& 65& 2& 4& 30& 60\\ 70-80& 7& 75& 3& 9& 21& 63\\ & 150& & & & -6& 366\\ \hline\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}\\ \text{\hspace{0.17em}}=45+\frac{-6}{150}×10\\ \text{\hspace{0.17em}}=45-0.4\text{\hspace{0.17em}}=44.6\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\left(10\right)}^{2}}{{\left(150\right)}^{2}}\left[150×366-{\left(-6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}\left(54900-36\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}×54864\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=243.84\\ \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{243.84}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15.61\end{array}$

Since, both groups have equal mean. Group B has greater standard deviation. So, group B is more variable than group A.

Q.14 From the prices of shares X and Y below, find out which is more stable in value:

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

Ans

For prices of share X:

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{35+54+52+53+56+58+52+50+51+49}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{510}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=51\end{array}$ $\begin{array}{l}\begin{array}{|ccc|}\hline {\mathrm{x}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 35& 35-51=-16& 256\\ 54& 54-51=3& 9\\ 52& 52-51=1& 1\\ 53& 53-51=2& 4\\ 56& 56-51=5& 25\\ 58& 58-51=7& 49\\ 52& 52-51=1& 1\\ 50& 50-51=-1& 1\\ 51& 51-51=0& 0\\ 49& 49-51=-2& 4\\ & & 350\\ \hline\end{array}\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{N}}\\ =\frac{350}{10}\\ =35\\ \therefore \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ =\sqrt{35}\\ =5.91\\ \mathrm{C}.\mathrm{V}.\left(\mathrm{share}\text{X}\right)=\frac{\mathrm{\sigma }}{\overline{\mathrm{x}}}×100\\ =\frac{5.91}{51}×100\\ =11.59\left(\mathrm{Appox}.\right)\end{array}$

For prices of share Y:

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{y}}\right)=\frac{108+107+105+105+106+107+104+103+104+101}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1050}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=105\\ \begin{array}{|ccc|}\hline {\mathrm{y}}_{\mathrm{i}}& \left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)& {\left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)}^{2}\\ 108& 108-105=3& 9\\ 107& 107-105=2& 4\\ 105& 105-105=0& 0\\ 105& 105-105=0& 0\\ 106& 106-105=1& 1\\ 107& 107-105=2& 4\\ 104& 104-105=-1& 1\\ 103& 103-105=-2& 4\\ 104& 104-105=-1& 1\\ 101& 101-105=-4& 16\\ & & 40\\ \hline\end{array}\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{N}}\\ =\frac{40}{10}\\ =4\\ \therefore \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ =\sqrt{4}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\end{array}$ $\begin{array}{l}=2\\ \mathrm{C}.\mathrm{V}.\left(\mathrm{share}\text{Y}\right)=\frac{\mathrm{\sigma }}{\overline{\mathrm{y}}}×100\\ =\frac{2}{105}×100\\ =1.9<11.59\\ \mathrm{C}.\mathrm{V}.\text{of prices of share Y is lesser than that of share X.}\\ \text{Thus, the prices of share Y are more stable than that}\\ \text{of share X.}\end{array}$

Q.15 An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A Firm B
No. of wage earners 586 648
Mean of monthly wages Rs. 5253 Rs. 5253
Variance of distribution of wages 100 121

1. Which firm A or B pays larger amount as monthly wages?
2. Which firm, A or B, shows greater variability in individual wages?

Ans

$\begin{array}{l}\mathrm{i}.\text{ Monthly wage of firm A = Rs. 5253}\\ \text{ Number of workers in firm A = 586}\\ \text{ Total paid amount in wages = Rs. 5253 × 586}\\ \text{ = Rs. 3078258}\\ \text{ Monthly wage of firm B = Rs. 5253}\\ \text{ Number of workers in firm A = 648}\\ \text{ Total paid amount in wages = Rs. 5253 x 648}\\ \text{ = Rs.3403944}\\ \text{Thus, firm B pays the larger amount as monthly wages.}\\ \text{ii. Variance of distribution of wages in firm A = 100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Standard deviation of wages in firm A = }\sqrt{\text{100}}\\ \text{ = 10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}Variance of distribution of wages in firm B = 121}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Standard deviation of wages in firm B = }\sqrt{\text{121}}\\ \text{ = 11}\\ \text{Since the average monthly wages in both the firms is same,}\\ \text{i.e., Rs. 5253, therefore, the firm with greater standard deviation}\\ \text{will have more variability.}\\ \text{Thus, the firm B has greater variability in the individual wages.}\end{array}$

Q.16 The following is the record of goals scored by team A in a football session:

 No. of goals scored 0 1 2 3 4 No. of matched 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Ans

$\begin{array}{|ccccc|}\hline \begin{array}{l}No.\text{of goals}\\ \text{}\left({\text{x}}_{\text{i}}\right)\end{array}& \begin{array}{l}No.\text{of matches}\\ \text{}{\left(\text{f}\right)}_{\text{i}}\end{array}& {x}_{i}-\overline{x}& {\left({x}_{i}-\overline{x}\right)}^{2}& {f}_{i}{\left({x}_{i}-\overline{x}\right)}^{2}\\ 0& 1& 0-2=-2& 4& 4\\ 1& 9& 1-2=-1& 1& 9\\ 2& 7& 2-2=0& 0& 0\\ 3& 5& 3-2=1& 1& 5\\ 4& 3& 4-2=2& 4& 12\\ & 25& & & 30\\ \hline\end{array}$ $\begin{array}{l}\text{Mean = }\frac{\sum _{\text{i=1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}}{\text{n}}\\ \text{ = }\frac{\text{50}}{\text{25}}\\ \text{ = 2}\\ \text{Variation}\left({\text{σ}}^{\text{2}}\right)\\ \text{= }\frac{\sum _{\text{i=1}}^{\text{n}}{\text{f}}_{\text{i}}{\left({\text{x}}_{\text{i}}\text{–}\overline{\text{x}}\right)}^{\text{2}}}{\text{n}}\\ \text{= }\frac{\text{30}}{\text{25}}\\ \text{= 1.2}\\ \text{Standard deviation}\\ \text{= }\sqrt{\text{1.2}}\\ \text{= 1.09}\\ \text{Standard deviation of team B}\\ \text{= 1.25 goals}\\ \text{Mean of team B}\\ \text{= 2}\\ \text{The team with lower standard deviation is more consistent.}\\ \text{Therefore, team A is more consistent.}\end{array}$

Q.17 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

$\begin{array}{l}\sum _{i=1}^{50}{x}_{i}=212,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{x}_{i}}^{2}=902.8,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{y}_{i}=261,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{y}_{i}}^{2}=1457.6\\ \mathbf{Which}\mathbf{}\mathbf{is}\mathbf{}\mathbf{more}\mathbf{}\mathbf{varying}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{length}\mathbf{}\mathbf{or}\mathbf{}\mathbf{weight}\mathbf{?}\end{array}$

Ans

$\begin{array}{l}\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}=212,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{\mathrm{i}=1}^{50}{{\mathrm{x}}_{\mathrm{i}}}^{2}=902.8,\\ \mathrm{Mean}=\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{212}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4.24\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{{\mathrm{x}}_{\mathrm{i}}}^{2}}{50}-{\left(\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{902.8}{50}-{\left(\frac{212}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=18.056-17.978\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.078\\ \text{Standandard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{0.078}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.28\\ \mathrm{C}.\mathrm{V}.\text{of length}=\frac{\mathrm{\sigma }}{\overline{\mathrm{x}}}×100\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6.6\\ \text{\hspace{0.17em}}\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}=261,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{\mathrm{i}=1}^{50}{{\mathrm{y}}_{\mathrm{i}}}^{2}=1457.6\\ \mathrm{Mean}\left(\overline{\mathrm{y}}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{261}{50}\text{\hspace{0.17em}\hspace{0.17em}}=5.22\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{{\mathrm{y}}_{\mathrm{i}}}^{2}}{50}-{\left(\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1457.6}{50}-{\left(\frac{261}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=29.15-27.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.89\\ \text{Standandard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{1.89}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.37\\ \mathrm{C}.\mathrm{V}.\text{of weight}=\frac{\mathrm{\sigma }}{\overline{\mathrm{y}}}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1.37}{5.22}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=26.24\end{array}$ $\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}C.V. of weights is greater than the C.V. of lengths.}\\ \text{Therefore, weights are more varing than lengths.}\end{array}$

Q.18 The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Ans

Let remaining two observations be x and y.

$\begin{array}{l}\mathrm{Therefore},\text{the series is}6,\text{}7,\text{}10,\text{}12,\text{\hspace{0.17em}}12,\text{\hspace{0.17em}}13,\mathrm{x}\text{and y.}\\ \text{Then, Mean}=\frac{6+7+10+12+12+13+\mathrm{x}+\text{y}}{8}\\ ⇒9=\frac{60+\mathrm{x}+\text{y}}{8}\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}72=60+\mathrm{x}+\text{y}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=72-60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=12...\left(\mathrm{i}\right)\\ \mathrm{Variance}=\frac{\sum _{\mathrm{i}=1}^{8}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9.25=\frac{\left\{\begin{array}{l}{\left(6-9\right)}^{2}+{\left(7-9\right)}^{2}+{\left(10-9\right)}^{2}+{\left(12-9\right)}^{2}+\\ {\left(12-9\right)}^{2}+{\left(13-9\right)}^{2}+{\left(\mathrm{x}-9\right)}^{2}+{\left(\mathrm{y}-9\right)}^{2}\end{array}\right\}}{8}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}74=9+4+1+9+9+16+{\mathrm{x}}^{2}-18\mathrm{x}+81+{\mathrm{y}}^{2}-18\mathrm{y}+81\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}74=210+{\mathrm{x}}^{2}-18\left(\mathrm{x}+\mathrm{y}\right)+{\mathrm{y}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}74=210+{\mathrm{x}}^{2}-18\left(12\right)+{\mathrm{y}}^{2}\left[\mathrm{From}\text{eqation}\left(\mathrm{i}\right)\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}74=210+{\mathrm{x}}^{2}-216+{\mathrm{y}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}80={\mathrm{x}}^{2}+{\mathrm{y}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=80...\left(\mathrm{ii}\right)\end{array}$ $\mathrm{Squarring}\text{equation}\left(\mathrm{i}\right),\text{we get}$ $\text{}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}+\mathrm{y}\right)}^{2}={\left(12\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{x}}^{2}+2\mathrm{xy}+{\mathrm{y}}^{2}=144\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{xy}=144-\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{xy}=144-80\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{xy}=64...\left(\mathrm{iii}\right)\\ \mathrm{Now},\\ {\left(\mathrm{x}-\mathrm{y}\right)}^{2}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}-2\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=80-64\\ ⇒{\left(\mathrm{x}-\mathrm{y}\right)}^{2}=16\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=±4...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation\hspace{0.17em}}\left(\mathrm{i}\right)\text{and}\left(\mathrm{iv}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=8\text{and y}=\text{4}\left(\mathrm{By}\text{taking, x}-\text{y}=\text{4}\right)\\ \mathrm{or}\text{\hspace{0.17em}}\mathrm{x}=4\text{and y}=\text{8}\left(\mathrm{By}\text{taking, x}-\text{y}=-\text{4}\right)\\ \mathrm{Thus},\text{the remaining two observations are 4 and 8.}\end{array}$

Q.19 The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Ans

Let remaining two observations be x and y.

$\begin{array}{l}\mathrm{Therefore},\text{the series is}2,\text{}4,\text{}10,\text{}12,\text{\hspace{0.17em}}14,\mathrm{x}\text{and y.}\\ \text{Then, Mean}=\frac{2+4+10+12+14+\mathrm{x}+\text{y}}{7}\\ ⇒8=\frac{42+\mathrm{x}+\text{y}}{7}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}56=42+\mathrm{x}+\text{y}\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}}\mathrm{x}+\mathrm{y}=56-42\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=14...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}=\frac{\sum _{\mathrm{i}=1}^{8}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}16=\frac{\left\{\begin{array}{l}{\left(2-8\right)}^{2}+{\left(4-8\right)}^{2}+{\left(10-8\right)}^{2}+{\left(12-8\right)}^{2}+\\ {\left(14-8\right)}^{2}+{\left(\mathrm{x}-8\right)}^{2}+{\left(\mathrm{y}-8\right)}^{2}\end{array}\right\}}{7}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}112=36+16+4+16+36+{\mathrm{x}}^{2}-16\mathrm{x}+64+{\mathrm{y}}^{2}-16\mathrm{y}+64\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}112=236+{\mathrm{x}}^{2}-16\left(\mathrm{x}+\mathrm{y}\right)+{\mathrm{y}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}112=236+{\mathrm{x}}^{2}-16\left(14\right)+{\mathrm{y}}^{2}\left[\mathrm{From}\text{eqation}\left(\mathrm{i}\right)\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}112=236+{\mathrm{x}}^{2}-224+{\mathrm{y}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}100={\mathrm{x}}^{2}+{\mathrm{y}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=100...\left(\mathrm{ii}\right)\\ \mathrm{Squarring}\text{equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}+\mathrm{y}\right)}^{2}={\left(14\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{x}}^{2}+2\mathrm{xy}+{\mathrm{y}}^{2}=196\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}100+2\mathrm{xy}=196\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{xy}=96...\left(\mathrm{iii}\right)\\ \mathrm{Now},\\ {\left(\mathrm{x}-\mathrm{y}\right)}^{2}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}-2\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100-96\\ ⇒{\left(\mathrm{x}-\mathrm{y}\right)}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=±2...\left(\mathrm{iv}\right)\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{From}\text{equation\hspace{0.17em}}\left(\mathrm{i}\right)\text{and}\left(\mathrm{iv}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=8\text{and y}=\text{6}\left(\mathrm{By}\text{taking, x}-\text{y}=\text{2}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}x}=6\text{and y}=\text{8}\left(\mathrm{By}\text{taking, x}-\text{y}=-\text{2}\right)\\ \mathrm{Thus},\text{the remaining two observations are 6 and 8.}\end{array}$

Q.20 The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Mean}\text{of 6 observations}=8\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Sum}\text{of 6 observations}=6×8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=48\\ \mathrm{When}\text{each observation is multiplied by 3,}\\ \text{New mean}=\left(\frac{\sum 3\mathrm{x}}{6}\right)=\frac{3×48}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24\\ \text{Standard deviation}=4\\ \mathrm{Variance}\text{of 6 observations\hspace{0.17em}}=16\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{6}-{\left(\frac{\sum \mathrm{x}}{6}\right)}^{2}=16\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{6}-{\left(8\right)}^{2}=16\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{6}=16+64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sum {\mathrm{x}}^{2}=6×80\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=480\\ \mathrm{When}\text{each observation is multiplied by 3,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\sum {\left(3\mathrm{x}\right)}^{2}=9\sum {\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9×480\\ \mathrm{New}\text{variance of 6 observations}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sum {\left(3\mathrm{x}\right)}^{2}}{6}-{\left(\frac{\sum 3\mathrm{x}}{6}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9×480}{6}-{\left(24\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=720-576\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=144\\ \mathrm{New}\text{standard deviation}=\sqrt{144}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \mathrm{Thus},\text{new mean and new standard deviatiion of 6 observations}\\ \text{are 24 and 12 respectively.}\end{array}$

Q.21

$\begin{array}{l}\mathbf{\text{Given that}}\overline{\mathbf{\text{x}}}{\mathbf{\text{is the mean and σ}}}^{\mathbf{\text{2}}}\mathbf{\text{is the variance of n}}\\ {\mathbf{\text{observations x}}}_{\mathbf{\text{1}}}{\mathbf{\text{, x}}}_{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}{\mathbf{\text{,x}}}_{\mathbf{\text{n}}}\mathbf{\text{.\hspace{0.17em}Prove that the mean and}}\\ {\mathbf{\text{variance of the observations ax}}}_{\mathbf{\text{1}}}{\mathbf{\text{, ax}}}_{\mathbf{\text{2}}}{\mathbf{\text{, ax}}}_{\mathbf{\text{3}}}\mathbf{\text{,}}\mathbf{.}\mathbf{\dots }{\mathbf{\text{, ax}}}_{\mathbf{\text{n}}}\mathbf{\text{are\hspace{0.17em}a}}\overline{\mathbf{\text{x}}}\mathbf{\text{}}\\ {\mathbf{\text{and a}}}^{\mathbf{\text{2}}}{\mathbf{\text{σ}}}^{\mathbf{\text{2}}}\mathbf{\text{, respectively, (a¹0).}}\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mean of n observations = }\overline{\text{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+x}}_{\text{3}}\text{+}...{\text{+x}}_{\text{n}}}{\text{n}}\text{ = }\overline{\text{x}}\text{}...\left(\text{i}\right)\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Variance of n observations = σ}}^{\text{2}}\end{array}$ $\begin{array}{l}\frac{\left({{\text{x}}_{\text{1}}}^{\text{2}}{{\text{+x}}_{\text{2}}}^{\text{2}}{{\text{+x}}_{\text{3}}}^{\text{2}}\text{+}...{{\text{+x}}_{\text{n}}}^{\text{2}}\right)}{\text{n}}\text{–}{\left(\frac{{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+x}}_{\text{3}}\text{+}...{\text{+x}}_{\text{n}}}{\text{n}}\right)}^{\text{2}}{\text{ = σ}}^{\text{2}}\\ \frac{\left({{\text{x}}_{\text{1}}}^{\text{2}}{{\text{+x}}_{\text{2}}}^{\text{2}}{{\text{+x}}_{\text{3}}}^{\text{2}}\text{+}...{{\text{+x}}_{\text{n}}}^{\text{2}}\right)}{\text{n}}\text{–}{\left(\overline{\text{x}}\right)}^{\text{2}}{\text{= σ}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}\frac{\left({{\text{x}}_{\text{1}}}^{\text{2}}{{\text{+x}}_{\text{2}}}^{\text{2}}{{\text{+x}}_{\text{3}}}^{\text{2}}\text{+}...{{\text{+x}}_{\text{n}}}^{\text{2}}\right)}{\text{n}}{\text{= σ}}^{\text{2}}\text{+}{\left(\overline{\text{x}}\right)}^{\text{2}}\text{}...\left(\text{ii}\right)\\ {\text{Mean of ax}}_{\text{1}}{\text{, ax}}_{\text{2}}{\text{, ax}}_{\text{3}}\text{,}.\dots {\text{, ax}}_{\text{n}}\text{=}\frac{{\text{ax}}_{\text{1}}{\text{+ ax}}_{\text{2}}{\text{+ ax}}_{\text{3}}\text{+}.\dots {\text{+ ax}}_{\text{n}}}{\text{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\frac{\text{a}\left({\text{x}}_{\text{1}}{\text{+ x}}_{\text{2}}{\text{+ x}}_{\text{3}}\text{+}.\dots {\text{+ x}}_{\text{n}}\right)}{\text{n}}\\ {\text{Mean of ax}}_{\text{1}}{\text{, ax}}_{\text{2}}{\text{, ax}}_{\text{3}}\text{,}.\dots {\text{, ax}}_{\text{n}}\text{= a}\overline{\text{x}}\\ {\text{Variance of ax}}_{\text{1}}{\text{, ax}}_{\text{2}}{\text{, ax}}_{\text{3}}\text{,}.\dots {\text{, ax}}_{\text{n}}\\ \text{= }\frac{\left({\text{a}}^{\text{2}}{{\text{x}}_{\text{1}}}^{\text{2}}{\text{+a}}^{\text{2}}{{\text{x}}_{\text{2}}}^{\text{2}}{\text{+a}}^{\text{2}}{{\text{x}}_{\text{3}}}^{\text{2}}\text{+}...{\text{+a}}^{\text{2}}{{\text{x}}_{\text{n}}}^{\text{2}}\right)}{\text{n}}\text{ - }{\left(\frac{{\text{ax}}_{\text{1}}{\text{+ax}}_{\text{2}}{\text{+ax}}_{\text{3}}\text{+}...{\text{+ax}}_{\text{n}}}{\text{n}}\right)}^{\text{2}}\\ {\text{= a}}^{\text{2}}\frac{\left({{\text{x}}_{\text{1}}}^{\text{2}}{{\text{+x}}_{\text{2}}}^{\text{2}}{{\text{+x}}_{\text{3}}}^{\text{2}}\text{+}...{{\text{+x}}_{\text{n}}}^{\text{2}}\right)}{\text{n}}{\text{ - a}}^{\text{2}}{\left(\frac{{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+x}}_{\text{3}}\text{+}...{\text{+x}}_{\text{n}}}{\text{n}}\right)}^{\text{2}}\\ {\text{= a}}^{\text{2}}\left\{{\text{σ}}^{\text{2}}\text{+}{\left(\overline{\text{x}}\right)}^{\text{2}}\right\}{\text{-a}}^{\text{2}}{\left(\overline{\text{x}}\right)}^{\text{2}}\\ {\text{= a}}^{\text{2}}{\text{σ}}^{\text{2}}{\text{+a}}^{\text{2}}{\left(\overline{\text{x}}\right)}^{\text{2}}{\text{-a}}^{\text{2}}{\left(\overline{\text{x}}\right)}^{\text{2}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}=}{\mathrm{a}}^{2}{\mathrm{\sigma }}^{2}\\ \mathrm{Thus},\text{the mean and the variance of}{\mathrm{ax}}_{1},{\mathrm{ax}}_{2},{\mathrm{ax}}_{3},.\dots ,{\mathrm{ax}}_{\mathrm{n}}\text{}\\ \text{are a}\overline{\text{x}}{\text{and a}}^{\text{2}}{\mathrm{\sigma }}^{\text{2}}\text{.}\end{array}$

Q.22 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.
(ii) If it is replaced by 12.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Mean}\text{value of 20 observations}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{sum of 20 observations}=200\\ \text{Standard deviation of 20 observations}=2\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\text{of 20 observations}=4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{20}-{\left(\frac{\sum \mathrm{x}}{20}\right)}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{20}-{\left(10\right)}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{20}=100+4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{}\sum {\mathrm{x}}^{2}=104×20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Incorrect}\text{observation}=8\\ \left(\mathrm{i}\right)\mathrm{If}\text{wrong item is omitted, then}\\ \text{Remaining observations}=20-1\\ \text{}=19\end{array}$ $\begin{array}{l}\text{New mean of remaining observations}=\frac{200-8}{19}=\frac{192}{19}\\ \text{}=10.1\\ \mathrm{New}\text{variance of remaining observations}\\ \text{}=\frac{\sum {\mathrm{x}}^{2}-{8}^{2}}{19}-{\left(\frac{\sum \mathrm{x}-8}{19}\right)}^{2}\\ \text{}=\frac{104×20-{8}^{2}}{19}-{\left(\frac{200-8}{19}\right)}^{2}\\ \text{}=\frac{2016}{19}-{\left(10.1\right)}^{2}\\ \text{}=106.1-102.01\\ \text{}=4.09\\ \mathrm{New}\text{standard deviation of remaining observations}\\ \text{}=\sqrt{4.09}\\ \text{}=2.02\\ \left(\mathrm{ii}\right)\mathrm{When}\text{8 is replaced by 12,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}New mean of 20 observations}=\frac{200-8+12}{20}\\ \text{}=\frac{204}{20}\\ \text{}=10.2\\ \mathrm{New}\text{variance of 20 observations}\\ \text{}=\frac{\sum {\mathrm{x}}^{2}-{8}^{2}+{12}^{2}}{20}-{\left(\frac{\sum \mathrm{x}-8+12}{20}\right)}^{2}\\ \mathrm{New}\text{variance of 20 observations}=\frac{104×20-{8}^{2}+{12}^{2}}{20}-{\left(\frac{200-8+12}{20}\right)}^{2}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{2080-64+144}{20}-{\left(10.2\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2160}{20}-104.04\\ \text{\hspace{0.17em}\hspace{0.17em}}=108-104.04\\ \mathrm{New}\text{variance of 20 observations}=3.96\\ \mathrm{New}\text{standard deviation}=1.9899\\ \text{\hspace{0.17em}\hspace{0.17em}}=1.99\text{\hspace{0.17em}}\left(\mathrm{Approx}\right)\end{array}$

Q.23 The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard 12 15 20
Deviation
Which of the three subjects shows the highest variability in marks and which shows the lowest?

Ans

$\begin{array}{l}\mathrm{C}.\mathrm{V}.\text{of marks in Mathematics}=\frac{\mathrm{\sigma }}{\mathrm{Mean}}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{12}{42}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=28.57\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}.\mathrm{V}.\text{of marks in Physics}=\frac{\mathrm{\sigma }}{\mathrm{Mean}}×100\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{32}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=46.87\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}.\mathrm{V}.\text{of marks in Chemistry}=\frac{\mathrm{\sigma }}{\mathrm{Mean}}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{20}{40.9}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=48.89\\ \mathrm{Since},\text{C.V. of Chemistry is the greatest and C. V. of}\\ \text{Mathematics is the lowest.So, highest variability is in}\\ \text{the marks of Chemistry and lowest variability is in the}\\ \text{marks of Mathematics.}\end{array}$

Q.24 The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Ans

$\begin{array}{l}\mathrm{Mean}\text{value of 100 observations}=20\\ \mathrm{Total}\text{sum of 100 observations}=2000\\ \text{Standard deviation of 100 observations}=3\\ \mathrm{Variance}\text{of 100 observations}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{100}-{\left(\frac{\sum \mathrm{x}}{100}\right)}^{2}=9\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{100}-{\left(20\right)}^{2}=9\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sum {\mathrm{x}}^{2}}{100}=400+9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sum {\mathrm{x}}^{2}=409×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Incorrect}\text{observations}=21,\text{\hspace{0.17em}}21,\text{\hspace{0.17em}}18\\ \mathrm{If}\text{wrong observations are omitted, then}\\ \text{Remaining observations}=100-3\\ \text{}=97\\ \text{New mean of remaining observations}=\frac{2000-21-21-18}{97}\\ \text{}=\frac{1940}{97}\\ \text{}=20\\ \mathrm{New}\text{variance of remaining observations}\\ \text{}=\frac{\sum {\mathrm{x}}^{2}-{21}^{2}-{21}^{2}-{18}^{2}}{97}\\ -{\left(\frac{\sum \mathrm{x}-21-21-18}{97}\right)}^{2}\\ \text{}=\frac{40900-441-441-324}{97}-{\left(20\right)}^{2}\\ \text{}=\frac{39694}{97}-400\\ \text{}=409.216-400\\ \text{}=9.216\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{New}\text{standard deviation}=\sqrt{9.216}\\ \text{}=3.036\end{array}$

Thus, the new mean and new standard deviation are 20 and 3.036 respectively.

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### 1. What are the key formulas in Statistics Class 11?

Mean, median, variance, and standard deviation formulae are important in Class 11. You’ll be able to develop a clear knowledge of the topics contained in the Chapter 15 Class 11 Mathematics once you’ve completed them. To do so, use the examples and activities in the NCERT Class 11 Mathematics CBSE book to solve and practise each sum.

### 2. Explain the concept of standard deviation, which is covered in Chapter 15 of the NCERT Solutions for Class 11 Mathematics.

The measurement of variation or deviation of a given set of values is dealt with by standard deviation. The range is determined by the standard deviation level. Students should obtain the solutions accessible at Extramarks to better grasp this term. Based on the needs of the students, they are presented in chapter and exercise format.

### 3. From where and how can I get the NCERT solutions for CBSE Mathematics Class 11 Chapter 15 Statistics?

The NCERT Solutions for Class 11 Chapter 15, which can be found in this post, is created by subject experts and professionals affiliated with Extramarks who have considerable experience in the field of teaching and are well-versed in the entire course. The solutions are specifically created with NCERT recommendations in mind. Because they are written by subject-matter experts, these answers and notes save time and are reliable for preparation.

### 4. How much should you practise Mathematics for NCERT Class 11?

While there is no set time for practising equations, the time required varies from person to person. Students are advised to select and practise challenging sums and equations from the Class 11 Mathematics NCERT Solutions Chapter 15. It’s also vital to set aside a few hours each week to practise those difficult equations, since this will help you study in the most efficient way possible.

### 5. Do I have to practise all of the questions in NCERT Solutions Class 11 Mathematics Statistics?

Constant practise is the key to mastering Ch 15 Mathematics Class 11 on Statistics. Students must memorise a number of formulas and conduct extensive calculations. A  simple error while solving these problems can cost the entire mark of that question. As a result, students should review these answers on a frequent basis to keep their skills sharp.

### 6. What is the total number of questions in NCERT Solutions Class 11 Mathematics Chapter 15 Statistics?

In Chapter 15 Class 11 Mathematics, there are 34 problems divided among three exercises and one miscellaneous exercise. These questions, handpicked by Mathematics professionals, provide children with a 360-degree picture of the entire chapter. They range from simple formula-based to complex word problems.