# NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

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NCERT Solutions Class 11 by Extramarks

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## H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

One can access the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 from the Extramarks official website or mobile application. It is available every time one downloads it, as per their convenience.H3 – NCERT Solutions for Class 11 Maths Chapters

Extramarks provides NCERT Solutions for all chapters of Class 11 Mathematics in both Hindi and English. The NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2 in English are available on this page.Below are also the links to the different chapters of Class 1 Mathematics.

NCERT Solutions for Class 11 Mathematics Chapters

• Chapter 1 – Sets
• Chapter 2 – Relations and Functions
• Chapter 3 – Trigonometric Functions
• Chapter 4 – Principle of Mathematical Induction
• Chapter 5 – Complex Numbers and Quadratic Equations
• Chapter 6 – Linear Inequalities
• Chapter 7 – Permutations and Combinations
• Chapter 8 – Binomial Theorem
• Chapter 9 – Sequences and Series
• Chapter 10 – Straight Lines
• Chapter 11 – Conic Sections
• Chapter 12 – Introduction to 3D Geometry
• Chapter 13 – Limits and Derivatives
• Chapter 14 – Mathematical Reasoning
• Chapter 15 – Statistics
• Chapter 16 – Probability

The different topics covered under the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 are given below:

1. Limitations of mean deviation
2. Variance and standard deviation
3. Standard deviation
4. Standard deviation of a discrete frequency distribution
5. Standard deviation of a continuous frequency distribution
6. Evaluating variance and standard deviation using the shortcut method

### H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

The NCERT Class 11 Chapter 15 has other exercises as well. Below is the breakdown of the different exercises and the number of questions in each one of them.

• Exercise 15.1 – 12 questions and their solutions
• Exercise 15.2 -10 questions and their solutions
• Exercise 15.3 -5 questions and their solutions
• Miscellaneous exercise – 7 questions and their solutions

### H3 – All Topics Under NCERT Class 11 Maths for Exercise 15.2

Class 11 Mathematics NCERT Solution Chapter 15 – Statistics (Term I)

Statistics is an important chapter in advanced Mathematics. Students will discover some dispersion metrics and computation techniques for both grouped and ungrouped data in this chapter. Measures of dispersion, mean deviation, range, variance, and standard deviation of grouped and ungrouped data, as well as the study of frequency distributions, are only a few of the various ideas covered in this chapter. Students can easily get a better score after they are familiar with the themes and strategies for answering the questions.

The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 deals with all these topics in a comprehensive yet easy-to-understand format.

### H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

These answers come with thorough explanations and examples drawn from actual scenarios to make them easier to recall. Students can look up NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2 and other solutions for reference.

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Q.1 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.

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

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3465}{10}-272.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=346.5-272.25\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)\text{\hspace{0.17em}}=74.25\end{array}$

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.2 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.3 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.

$\begin{array}{|ccccccc|}\hline \mathrm{Classes}& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{30}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 0-30& 2& 15& -3& 9& -6& 18\\ 30-60& 3& 45& -2& 4& -6& 12\\ 60-90& 5& 75& -1& 1& -5& 5\\ 90-120& 10& 105=\mathrm{A}& 0& 0& 0& 0\\ 120-150& 3& 135& 1& 1& 3& 3\\ 150-180& 5& 165& 2& 4& 10& 20\\ 180-210& 2& 195& 3& 9& 6& 18\\ \mathrm{Total}& 30& & & & 2& 76\\ \hline\end{array}$ $\begin{array}{l}\text{}\text{}\end{array}$

$\begin{array}{l}Mean=A+\frac{\sum _{i=1}^{n}{f}_{i}{u}_{i}}{n}×h\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=105+\frac{2}{30}×30\end{array}$

$\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=107\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(30\right)}^{2}}{{\left(30\right)}^{2}}\left[30×76-{\left(2\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=2280-4=2276\end{array}$

8.

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

$\begin{array}{l}=\frac{{\left(10\right)}^{2}}{{\left(50\right)}^{2}}\left[50×68-{\left(10\right)}^{2}\right]\\ =\frac{1}{25}\left[3400-100\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=132\end{array}$

Q.4 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.

$\begin{array}{|ccccccc|}\hline Classes& {f}_{i}& {x}_{i}& {u}_{i}=\frac{{x}_{i}-A}{5}& {u}_{i}{}^{2}& {f}_{i}{u}_{i}& {f}_{i}{u}_{i}{}^{2}\\ 70-75& 3& 72.5& -4& 16& -12& 48\\ 75-80& 4& 77.5& -3& 9& -12& 36\\ 80-85& 7& 82.5& -2& 4& -14& 28\\ 85-90& 7& 87.5& -1& 1& -7& 7\\ 90-95& 15& 92.5=A& 0& 0& 0& 0\\ 95-100& 9& 97.5& 1& 1& 9& 9\\ 100-105& 6& 102.5& 2& 4& 12& 24\\ 105-110& 6& 107.5& 3& 9& 18& 54\\ 110-115& 3& 112.5& 4& 16& 12& 78\\ Total& 60& & & & 6& 254\\ \hline\end{array}$

$\mathrm{Mean}=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}=92.5+\frac{6}{60}×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=92.5+0.5\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=93\\ \text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \text{\hspace{0.17em} \hspace{0.17em}}=\frac{{\left(5\right)}^{2}}{{\left(60\right)}^{2}}\left[60×254-{\left(6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{3600}\left[15240-36\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{3600}×15204\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=105.52\\ \text{Standard deviation}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\sqrt{105.52}\\ \text{\hspace{0.17em}\hspace{0.17em}}=10.27\end{array}$

Q.5 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.

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

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