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# NCERT Solutions for Class 11 Maths Chapter 13 - Statistics Miscellaneous Exercise

Last updated date: 10th Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 13 Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 13 Statistics, of the covers the fundamentals of data analysis and interpretation. In this chapter, methods for calculating averages (mean, median, and mode) and measures of distribution (standard deviation) are covered. These skills are necessary for making good decisions on a variety of topics.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 13 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 13 Statistics
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 13: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 13 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

This Chapter's Miscellaneous Exercise has useful tasks that help in applying statistical techniques to real-world situations. Students who practice Class 11 Maths NCERT Solutions can analyse carefully and get useful information from data. Recognising these fundamental principles of statistics is necessary for additional research and real-world applications across many different areas. Get the latest CBSE Class 11 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 13 Statistics

### Miscellaneous Exercise

1. The mean and variance of eight observations are ${\text{9}}$ and ${\text{9}}{\text{.25}}$ respectively. If six of the observations are ${\text{6,}}\,{\text{7,10,}}\,{\text{12,}}\,{\text{12}}$ and ${\text{13}}$, find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are $6,\,7,10,\,12,\,12,\,13,\,x,\,y$.

Therefore, from the mean of the data it can be obtained that,

$\overline x = \dfrac{{6 + 7 + \,10 + 12 + \,12 + 13 + x + y}}{8} = 9$

$\Rightarrow 60 + x + y = 72$

$\Rightarrow x + y = 12$ ......(i)

Again, from the variance of the data it can be obtained that,

${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 9.25$

$\therefore \dfrac{1}{8}\left[ {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2} - (2)(9)\left( {x + y} \right) + \left( {2{{\left( 9 \right)}^2}} \right)} \right] = 9.25$

$\Rightarrow \dfrac{1}{8}\left[ {9 + 4 + 9 + 16 + {x^2} + {y^2} - (18)\left( {12} \right) + 162} \right] = 9.25$   [By, using (i)]

$\Rightarrow \dfrac{1}{8}\left[ {{x^2} + {y^2} - 6} \right] = 9.25$

$\Rightarrow {x^2} + {y^2} = 80$ ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 144$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

$2xy = 64$ ......(iv)

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

${x^2} + {y^2} - 2xy = 16$

$\Rightarrow x - y = \pm 4$ ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 4$, when $x - y = 4$ and $x = 4$ and $y = 8$, when $x - y = - 4$. Henceforth, the remaining two observations are $4$ and $8$.

2. The mean and variance of ${\text{7}}$ observations are ${\text{8}}$ and ${\text{16}}$ respectively. If five of the observations are ${\text{2,}}\,{\text{4,10,}}\,{\text{12}}$ and ${\text{14}}$. Find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are $2,\,4,10,\,12,\,14,\,x,\,y$.

Therefore, from the mean of the data it can be obtained that,

$\overline x = \dfrac{{2 + 4 + \,10 + 12 + \,14 + x + y}}{7} = 8$

$\Rightarrow 42 + x + y = 56$

$\Rightarrow x + y = 14$ ......(i)

Again, from the variance of the data it can be obtained that,

${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 16$

$\therefore \dfrac{1}{7}\left[ {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2} - (2)(8)\left( {x + y} \right) + \left( {2{{\left( 8 \right)}^2}} \right)} \right] = 16$

$\Rightarrow \dfrac{1}{7}\left[ {36 + 16 + 4 + 16 + {x^2} + {y^2} - (16)\left( {14} \right) + 128} \right] = 16$   [By, using (i)]

$\Rightarrow \dfrac{1}{7}\left[ {{x^2} + {y^2} + 12} \right] = 16$

$\Rightarrow {x^2} + {y^2} = 100$ ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 196$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

$2xy = 96$ ......(iv)

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

${x^2} + {y^2} - 2xy = 4$

$\Rightarrow x - y = \pm 2$ ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 6$, when $x - y = 2$and $x = 6$ and $y = 8$, when $x - y = - 2$. Henceforth, the remaining two observations are $6$ and $8$.

3. The mean and standard deviation of six observations are ${\text{8}}$ and ${\text{4}}$ respectively. If each observation is multiplied by ${\text{3}}$, find the new mean and new standard deviation of the resulting observations.

Ans: Assume the observations to be ${x_1},\,{x_2},\,{x_3},\,{x_4},\,{x_5},\,{x_6}$.

Therefore, the mean of the data is,

$\overline x = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8$

$\Rightarrow \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8$ ......(i)

Observe that when each of the observation is multiplied with $3$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_1} = 3{x_1}$

$\Rightarrow {x_1} = \dfrac{1}{3}{y_1},\,\forall i = 1,\,2,\,3,\,....,\,6\& i \in {\mathbb{Z}^ + }$

Now, the mean of the new data is,

$\therefore \overline y = \dfrac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}$

$\Rightarrow \overline y = \dfrac{{3({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6})}}{6}$

$\Rightarrow \overline y = 3 \times 8$      [By using (i)]

$\Rightarrow \overline y = 24$

Therefore, the standard deviation of the data is,

$\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}} = 4$

$\Rightarrow {\left( 4 \right)^2} = \dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{6}$

$\Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} = 96$ ......(ii)

Again, it can be observed from equations (i) and (ii) that $\overline y = 3\overline x$ and $\overline x = \dfrac{{\overline y }}{3}$ and hence on substituting the values of ${x_i}$ and $\overline x$ in equation (ii) it can be clearly obtained as shown below,

$\therefore \sum\limits_{i = 1}^6 {{{\left( {\dfrac{{{y_i}}}{3} - \dfrac{{\overline y }}{3}} \right)}^2}} = 96$

$\Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_i} - \overline y } \right)}^2}} = 864$

Henceforth, the standard deviation of the new data can be calculated as shown below,

$\therefore \sigma = \sqrt {\dfrac{{864}}{6}} = \sqrt {144} = 12$

4. Given that $\overline {\text{x}}$ is the mean and $\sigma^{2}$ is the variance of ${\text{n}}$ observations ${{\text{x}}_{\text{1}}}{\text{,}}\,{{\text{x}}_{\text{2}}}{\text{,}}\,......{\text{,}}\,{{\text{x}}_{\text{n}}}$.Prove that the mean and variance of observations ${\text{a}}{{\text{x}}_{\text{1}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{2}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{3}}}{\text{, }}.....{\text{,}}\,{\text{a}}{{\text{x}}_{\text{n}}}$ are ${\text{a}}\overline {\text{x}}$ and ${{\text{a}}^{\text{2}}}{{\sigma}}^{\text{2}}$, respectively $\left( {{\text{a}} \ne {\text{0}}} \right)$.

Ans: Observe that the given observations are ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x$ and ${\sigma ^2}$ respectively.

$\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{y_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$......(i)

Observe that when each of the observation is multiplied with $a$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_i} = a{x_i}$

$\Rightarrow {x_i} = \dfrac{1}{a}{y_i},\,\forall i = 1,\,2,\,3,\,....,\,n\& i \in {\mathbb{Z}^ + }$

Now, the mean of the new data is,

$\therefore \overline y = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}} = \dfrac{1}{n}\sum\limits_{i = 1}^n {a{x_i}} = \dfrac{a}{n}\sum\limits_{i = 1}^n {{x_i}} = a\overline x \,\,\,\,\,\,\,\left[ {\because \overline x = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right]$

Again on substituting the values of ${x_i}$ and $\overline x$ in equation (i) it can be clearly obtained as shown below,

$\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {\dfrac{{{y_i}}}{a} - \dfrac{{\overline y }}{a}} \right)}^2}} }}{n}$

$\Rightarrow {a^2}{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{y_i} - \overline y } \right)}^2}} }}{n}$

Henceforth, it can be clearly proved that the mean and variance of the new data is $a\overline x$ and ${a^2}{\sigma ^2}$, respectively.

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

(i) If wrong item is omitted

Ans: Observe that the incorrect number of observations, incorrect mean and the incorrect standard deviation are $20$, $10$ and $2$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x$ and ${\sigma ^2}$ respectively.

$\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n} = 10$

$\Rightarrow \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 10$

$\Rightarrow \sum\limits_{i = 1}^{20} {{x_i}} = 200$

Now, the correct mean is,

$\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{19} {{x_i}} }}{n}$

$\Rightarrow \overline x = \dfrac{{192}}{{19}} = 10.1\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{19} {{x_i}} = 192} \right]$

Again observe as shown below,

$\therefore \sigma = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} } = 2$

$\Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2}$

Therefore, the correct standard deviation of the data is,

$\sum\limits_{i = 1}^{20} {{x_i}^2} - {\left( 8 \right)^2}$

$\Rightarrow 2080 - 64$

$\Rightarrow 2016$

$\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{19} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow \sigma = \sqrt {\dfrac{{2016}}{{19}} - {{\left( {10.1} \right)}^2}}$

$\Rightarrow \sigma = \sqrt {1061.1 - 102.1} = \sqrt {4.09} = 2.02$

(ii) If it is replaced by ${\text{12}}$.

Ans: Observe that the incorrect sum of observations is $200$.

$\therefore \sum\limits_{i = 1}^{20} {{x_i}} = 200 - 8 + 12 = 204$

Now, the correct mean is,

$\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n}$

$\Rightarrow \overline x = \dfrac{{204}}{{20}} = 10.2\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{20} {{x_i}} = 204} \right]$

Again observe as shown below,

$\therefore \sigma = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} } = 2$

$\Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2}$

Therefore, the correct standard deviation of the data is,

$\sum\limits_{i = 1}^{20} {{x_i}^2} - {\left( 8 \right)^2} + {\left( {12} \right)^2}$

$\Rightarrow 2080 - 64 + 144$

$\Rightarrow 2160$

$\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow \sigma = \sqrt {\dfrac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}}$

$\Rightarrow \sigma = \sqrt {108 - 104.04} = \sqrt {3.96} = 1.98$

6. The mean and standard deviation of a group of ${\text{100}}$ observations were found to be ${\text{20}}$ and ${\text{3}}$, respectively. Later on it was found that ${\text{3}}$ observations are incorrect which were recorded as ${\text{21,}}\,{\text{21}}$ and ${\text{18}}$. Find the mean and standard deviation if the incorrect  observations are omitted.

Ans: Observe that the number of observations, incorrect mean and the incorrect standard deviation are $100$, $20$ and $3$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x$ and ${\sigma ^2}$ respectively.

$\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{n} = 20$

$\Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{{100}} = 20$

$\Rightarrow \sum\limits_{i = 1}^{100} {{x_i}} = 2000$

Now, the correct mean is,

$\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{97} {{x_i}} }}{n}$

$\Rightarrow \overline x = \dfrac{{2000 - 60}}{{97}} = 20\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{97} {{x_i}} = 1940} \right]$

Again observe as shown below,

$\therefore \sigma = \sqrt {\dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} } = 3$

$\Rightarrow 9 = \dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow 40900 = \sum\limits_{i = 1}^{100} {{x_i}^2}$

Therefore, the correct standard deviation of the data is,

$\sum\limits_{i = 1}^{100} {{x_i}^2} - (2){\left( {21} \right)^2} - {(18)^2}$

$\Rightarrow 40900 - 1206$

$\Rightarrow 39694$

$\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{97} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}}$

$\Rightarrow \sigma = \sqrt {\dfrac{{39694}}{{97}} - {{\left( {20} \right)}^2}}$

$\Rightarrow \sigma = \sqrt {409.22 - 400} = \sqrt {9.22} = 3.04$.

## Conclusion

NCERT Solutions for Statistics Class 11 Miscellaneous Exercise, are crucial for understanding how to work with data effectively. This chapter teaches methods like finding averages (mean, median, mode) and measures of spread (standard deviation) to analyze data accurately. It's important to focus on practicing these methods to become proficient in interpreting data and making informed decisions. By mastering these statistical techniques, students develop practical skills that are applicable across various fields, from business analytics to scientific research.

## Class 11 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 12 Questions & Solutions Exercise 13.2 10 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 13 - Statistics Miscellaneous Exercise

1. What is statistics in simple terms from Statistics Class 11 Miscellaneous Exercise?

Statistics Class 11 Miscellaneous Exercise involves collecting, analyzing, interpreting, and presenting data to understand patterns and trends. It helps in making informed decisions based on facts rather than assumptions. For example, statistics can be used to analyze survey results, predict future trends, or evaluate the effectiveness of a new product in the market.

2. Why is statistics important in everyday life in Class 11 Maths Statistics Miscellaneous Solutions?

Statistics plays a crucial role in everyday life by providing tools to analyze data from various fields such as economics, healthcare, and social sciences. It helps in understanding trends, making predictions, and solving real-world problems. For instance, statistical analysis can be used to study consumer behavior, assess risks in insurance, or track changes in climate patterns.

3. What are measures of central tendency in Class 11 Maths Statistics Miscellaneous Solutions?

Measures of central tendency, like mean, median, and mode, summarize the central or typical values in a data set. The mean is the average value, the median is the middle value when data is sorted, and the mode is the most frequently occurring value. These measures provide insights into the typical characteristics of a data set.

4. How do you calculate the mean of a data set in Class 11 Maths Statistics Miscellaneous Solutions?

To find the mean from Class 11 Maths Statistics Miscellaneous Solutions, add up all the values in the data set and divide by the number of values. It gives a single numerical value that represents the average or central value of the data. For example, if you have test scores of 80, 85, and 90, the mean score would be (80 + 85 + 90) / 3 = 85.

5. What is the median of a data set and how is it found in Miscellaneous Exercise Class 11 Chapter 13?

Miscellaneous Exercise Class 11 Chapter 13, the median is the middle value in a sorted list of numbers. If there is an odd number of data points, it's the middle number. If there is an even number, it's the average of the two middle numbers. For instance, in the data set {5, 7, 12, 15, 20}, the median is 12.

6. When should you use the mode in Miscellaneous Exercise Class 11 Chapter 13?

The mode is used to identify the most frequently occurring value or values in a data set. It's particularly useful for categorical data or when analyzing data with distinct peaks or clusters. For example, in survey data where respondents choose their favorite color, the mode indicates the most popular choice.

7. What does standard deviation measure in Miscellaneous Exercise Class 11 Chapter 13?

Class 11 Maths Chapter 13 Miscellaneous Solutions, standard deviation measures the dispersion or spread of data points from the mean. It indicates how much the values in a data set deviate from the average. A smaller standard deviation suggests that data points are close to the mean, while a larger standard deviation indicates greater variability.

8. How can statistics help in making predictions from Class 11 Maths Chapter 13 Miscellaneous Solutions?

Class 11 Maths Chapter 13 Miscellaneous Solutions, statistics uses data analysis to identify patterns and trends, which can be used to make predictions about future events or outcomes. By studying historical data, statisticians can estimate probabilities and forecast potential scenarios. For example, in weather forecasting, statistics is used to predict the likelihood of rain based on past weather patterns.

9. What are some common types of graphs used in Class 11 Maths Chapter 13 Miscellaneous Solutions?

Common types of graphs include bar graphs, pie charts, histograms, and line graphs. These graphs visually represent data and help in understanding relationships, trends, and distributions. For instance, a histogram displays the frequency distribution of numerical data, making it easier to interpret data patterns.

10. How do you interpret a histogram in Ch 13 Miscellaneous Class 11?

A histogram is a graphical representation of data distribution, where data is grouped into intervals or bins along the x-axis and the frequency of each interval is represented on the y-axis. It helps in visualizing the shape, center, and spread of data. A taller histogram bar indicates a higher frequency of data points within that interval.

11. Why is it important to understand probability in Ch 13 Miscellaneous Class 11?

Probability is essential in statistics as it quantifies uncertainty and measures the likelihood of events occurring based on data analysis. It helps in making informed decisions under uncertainty, such as predicting outcomes in games of chance or assessing risks in financial investments.

12. How can NCERT Solutions Ch 13 Miscellaneous Class 11 help in understanding statistics better?

NCERT Solutions Ch 13 Miscellaneous Class 11, provide detailed explanations and step-by-step solutions to problems in Chapter 13. They offer clarity on statistical concepts and methods, helping students practice different types of questions. By using NCERT Solutions, students can strengthen their understanding of statistics and prepare effectively for exams.