**NCERT Solutions Class 11 Maths Chapter 15 Statistics** – Here are all the NCERT solutions for Class 11 Maths Chapter 15. This solution contains questions, answers, images, explanations of the complete chapter 15 titled Of Statistics taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 15 Statistics After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 15 Statistics in one place.

## NCERT Solutions Class 11 Maths Chapter 15 Statistics

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 15 Statistics , Maths, Class 11.

Class | 11 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 15 |

Chapter Name |
Statistics |

### NCERT Solutions Class 11 Maths chapter 15 Statistics

Class 11, Maths chapter 15, Statistics solutions are given below in PDF format. You can view them online or download PDF file for future use.

Did you find NCERT Solutions Class 11 Maths chapter 15 Statistics helpful? If yes, please comment below. Also please like, and share it with your friends!

### NCERT Solutions Class 11 Maths chapter 15 Statistics- Video

You can also watch the video solutions of NCERT Class11 Maths chapter 15 Statistics here.

Video – will be available soon.

If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.

### Download NCERT Solutions Class 11 Maths chapter 15 Statistics In PDF Format

You can also download here the **NCERT Solutions Class 11 Maths chapter 15 Statistics** in PDF format.

Click Here to download NCERT Solutions for Class 11 Maths chapter 15 Statistics

### Question & Answer

Q.1:Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17

Ans :The given data is 4, 7, 8, 9, 10, 12, 13, 17 Mean of the data, \(\overline{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10\) The deviations of the respective observations from the mean \(\overline{x}, \text { i.e. } x_{i}-\overline{x}\) are –6, – 3, –2, –1, 0, 2, 3, 7 The absolute values of the deviations, i.e. \(\left|x_{i}-\overline{x}\right|\), are 6, 3, 2, 1, 0, 2, 3, 7 The required mean deviation about the mean is \(\operatorname{M.D.}(\overline{x})=\frac{\sum_{i=1}^{8}\left|x_{i}-\overline{x}\right|}{8}=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}=3\)

Q.2:Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans :The given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Mean of the given data, \(\overline{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}=50\) The deviations of the respective observations from the mean \(\overline{x}, \text { i.e. } x_{i}-\overline{x}\) are –12, 20, –2, –10, –8, 5, 13, –4, 4, –6 The absolute values of the deviations, i.e. \(\left|x_{i}-\overline{x}\right|\) , are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6 The required mean deviation about the mean is \(\begin{aligned} \mathrm{M.D.}(\overline{x}) &=\frac{\sum_{i=1}^{10}\left|x_{i}-\overline{x}\right|}{10} \\ &=\frac{12+20+2+10+8+5+13+4+4+6}{10} \\ &=\frac{84}{10} \\ &=8.4 \end{aligned}\)

Q.3:Find the mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Ans :The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 Here, the numbers of observations are 12, which is even. Arranging the data in ascending order, we obtain 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 \(\mathrm{M}=\frac{\left(\frac{12}{2}\right)^{t h} \text { observation }+\left(\frac{12}{2}+1\right)^{t h} \text { observation }}{2}\) \(\begin{aligned} &=\frac{6^{\text { th }} \text { observation }+7^{\text { th }} \text { observation }}{2} \\ &=\frac{13+14}{2}=\frac{27}{2}=13.5 \end{aligned}\) \(\begin{array}{l}{\text { The deviations of the respective observations from the median, i.e. } x_{i}-\mathrm{M} \text { , }} \\ {\text { are }} \\ {-3.5,-2.5,-2.5,-1.5,-1.5,-0.5,-0.5,0.5,2.5,2.5,3.5,3.5,4.5} \\ {\text { The absolute values of the deviations, }\left|x_{i}-\mathrm{M}\right|_{\text { are }}}\end{array}\) 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5 The required mean deviation about the median is \(\begin{aligned} \mathrm{M.D.}(\mathrm{M}) &=\frac{\sum_{i=1}^{12}\left|x_{i}-\mathrm{M}\right|}{12} \\ &=\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12} \\ &=\frac{28}{12}=2.33 \end{aligned}\)

Q.4:Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Ans :The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Here, the number of observations is 10, which is even. Arranging the data in ascending order, we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \(\begin{aligned} \text { Median } M &=\frac{\left(\frac{10}{2}\right)^{t h} \text { observation }+\left(\frac{10}{2}+1\right)^{t h} \text { observation }}{2} \text { observation } \\ &=\frac{5^{t h} \text { observation }+6^{\prime h} \text { observation }}{2} \\ &=\frac{46+49}{2}=\frac{95}{2}=47.5 \end{aligned}\) \(\begin{array}{l}{\text { The deviations of the respective observations from the median, i.e. } x_{i}-\mathrm{M} \text { , }} \\ {\text { are }} \\ {-11.5,-5.5,-2.5,-1.5,-1.5,-1.5,3.5,5.5,12.5,24.5} \\ {\text { The absolute values of the deviations, }\left|x_{i}-\mathrm{M}\right|_{\text { are }}}\end{array}\) 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5 Thus, the required mean deviation about the median is \(\begin{aligned} \mathrm{MD} \cdot(\mathrm{M}) &=\frac{\sum_{i=1}^{10}\left|x_{i}-\mathrm{M}\right|}{10}=\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10} \\ &=\frac{70}{10}=7 \end{aligned}\)

Q.5:Find the mean deviation about the mean for the data \(\begin{array}{llllll}{x_{i}} & {5} & {10} & {15} & {20} & {25} \\ {f_{i}} & {7} & {4} & {6} & {3} & {5}\end{array}\)

Ans :\(\begin{array}{l}{\mathrm{N}=\sum_{i=1}^{5} \mathrm{f}_{\mathrm{i}}=25} \\ {\sum_{i=1}^{5} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=350} \\ {\therefore \overline{\mathrm{x}}=\frac{1}{\mathrm{N}} \sum_{\mathrm{i}=1}^{5} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{1}{25} \times 350=14} \\ {\therefore \mathrm{MD}(\overline{\mathrm{x}})=\frac{1}{\mathrm{N}} \sum_{i=1}^{5} \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|=\frac{1}{25} \times 158=6.32}\end{array}\)

## NCERT / CBSE Book for Class 11 Maths

You can download the NCERT Book for Class 11 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 11 Maths
- Click here to buy NCERT Book for Class 11 Maths

### All NCERT Solutions Class 11

- NCERT Solutions for Class 11 Accountancy
- NCERT Solutions for Class 11 Biology
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Economics
- NCERT Solutions for Class 11 History
- NCERT Solutions for Class 11 Geography
- NCERT Solutions for Class 11 Political Science
- NCERT Solutions for Class 11 Sociology
- NCERT Solutions for Class 11 Psychology
- NCERT Solutions for Class 11 English
- NCERT Solutions for Class 11 Hindi
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Business Studies
- NCERT Solutions for Class 11 Statistics

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

Class 7 | Class 8 | Class 9 |

Class 10 | Class 11 | Class 12 |

Download the NCERT Solutions app for quick access to NCERT Solutions Class 11 Maths Chapter 15 Statistics. It will help you stay updated with relevant study material to help you top your class!

Previous Next

To get fastest exam alerts and government job alerts in India, join our Telegram channel.