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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.

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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.

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics

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

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