case study questions class 11 maths statistics

CBSE Case Study Questions for Class 11 Maths Statistics Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Statistics  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Statistics PDF

• Chapter 12 Introduction to 3D Geometry Case Study Questions
• Chapter 13 Limits and Derivatives Case Study Questions
• Chapter 14 Mathematical Reasoning Case Study Questions
• Chapter 16 Probability Case Study Questions

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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Class 11th Applied Mathematics - Descriptive Statistics Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Applied Mathematics Subject - Descriptive Statistics, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Descriptive statistics case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

Applied Mathematics

(ii) What is the central angle formed by the region represented by Italy i.e. , 30.3%

(iii) What is the total population in Germany is 5,54,40,000, then how many persons are infected by Corona Virus ?

(iv) What is the central angle formed by the region represented by Germany i.e., 10.2%

(v) If other region is having total population 1,50,00,000, then what is the population infected in Finland ?

(ii) What is the difference between the highest production item and the lowest production item ?

(iii) Ratio of production of Bananas and Potatoes is

(iv) If production of Pineapple and apples are equal and the sum of their production is equal to thrice the production of Pears, then which of the following equation represent this:

(v) Which two items have same production during the year?

(ii) Mean of the given data is

(iii) What is the mean deviation about the median of the given scores ?

(iv) If the scores 38 and 34 are replaced by 68 and 74 what will be the mean of the data ?

(v) Difference between maximum value of data and minimum vale of data is called

(ii) What is the variance of the given data ?

(iii) Relation between Variance and Standard Deviation is

(iv) What is the Standard Deviation of the data ?

(v) Mean of first 20 natural numbers is

(i) The value  \(\Sigma d\)   is 

(ii) What is the value  \(\Sigma d^{2}\)  is 

(iii) What is the value of n 2 in the given data?

(iv) What is rank correlation of the given data?

(v) What does very low value of the rank correlation in the above answer indicates?

In class of Statistics, teacher was discussing the concept of Measures of Correlation, in which he was discussing about Karl Pearson’s Coefficient of Correlation. During his class, he discussed the following few points on this: This is the best method for finding correlation between two variables provided the relationship between the two variables is linear. This method is also known as product-moment correlation coefficient. Pearson's correlation coefficient may be defined as the ratio of covariance between the two variables to the product of the standard deviations of the two variables. If the two variables are denoted by x and y and of the corresponding bivariates data are ( x i , y i) for i = 1, 2, 3, ...., n, then the coefficent of correlation between x and y due to Karl Pearson, is given by: r = r xy  \(\text { or } \quad r_{x y}=\frac{\operatorname{Cov}(x, y)}{\sqrt{\operatorname{Var} x} \cdot \sqrt{\operatorname{Var} y}}\)   =  \(\frac{\operatorname{Cov}(x, y)}{\sigma_{x} \cdot \sigma_{y}}\)   where,    \(\operatorname{Cov}(x, y)=\frac{\Sigma(x-\bar{x})(y-\bar{y})}{N}\)   =  \(\frac{\Sigma x y}{N}-\bar{x} \cdot \bar{y}\)   =  \(\sigma_{x}=\sqrt{\frac{\Sigma(x-\bar{x})^{2}}{N}}=\sqrt{\frac{\Sigma x^{2}}{N}-x^{-2}}\)   =  \(\sigma_{y}=\sqrt{\frac{\Sigma(y-\bar{y})^{2}}{N}}=\sqrt{\frac{\Sigma y^{2}}{N}-y^{-2}}\)   \(\text { If } x-\bar{x}, y-\bar{y}\)  are small fractions, we use  \(r=\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^{2}} \sqrt{\Sigma(y-\bar{y})^{2}}}\)  

If x, y are small numbers, we use  \(r=\frac{\Sigma x y-\frac{1}{N} \Sigma x \Sigma y}{\sqrt{\Sigma x^{2}-\frac{1}{N}(\Sigma x)^{2}} \sqrt{\sum y^{2}-\frac{1}{N}(\Sigma y)^{2}}}\)   For example: Find Karl Pearson's coefficient of correlation between X and Y for the following :

Following problem was given to students on the same concept: On the basis of the above information, answer the following questions: (i) Pearson’s correlation coefficient may be defined as the ratio of covariance between the two variables to the

(ii) What is the value  \(\Sigma x y\)   in this data:

(iii) What is the value \(\Sigma x^{2}\) ? 

(iv) What is the value of  \(\Sigma y^{2}\)   ?

(v) What is the value of Karl Pearson’s Coefficient of Correlation between x and y?

*****************************************

Descriptive statistics case study questions with answer key answer keys.

(i) (d):  45,45,000 Total population infected with Corona Virus \(=\frac{30.3}{100} \times 1,50,00,000\)   = 45,45,000 (ii) (d):  109.08° Central angle =  \(\frac{30.3}{100} \times 360^{\circ}\) = 109.08° (iii) (a):  56,54,880 Total population infected \(=\frac{10.2}{100} \times 55440000\) =  56,54,880 (iv) (c) :  37° Central angle  \(=\frac{10.2}{100} \times 360^{\circ}=36.7\) = 37° (v) (c):  6,30,000 Total population infected in Finland \(=\frac{4.2}{100} \times 1,50,00,000\) = 6,30,000

(i) (b):  potatoes Highest production item is potatoes i.e., 35,000 (ii) (d):  25000 Difference = 35000 – 10000 = 25000 (iii) (b): 5 : 7 \( \text { Ratio } =\frac{25000}{35000} \) \(=\frac{5}{7} \) = 5 : 7 (iv) (a): 2x = 3y Since production Pineapple and Apples are equal, let it be x and let production of Pears is y. Therefore,2x = 3y (v) (a):  Peas and Beans

(i) (c): 47 Arranging the data in ascending order = 34, 38, 42, 44, 46, 48, 54, 55, 63, 70 Median = A.M. of 5th and 6th observation \(=\frac{46+48}{2}=47\)   (ii) (b):   49.4 Sum of all the scores = 494 \(\text { Mean }=\frac{494}{10}=49.4\) (iii) (b):  8.6

\(\text { Mean Deviation }=\frac{1}{n} \times \sum\left|d_{i}\right|=\frac{86}{10}=8.6\) (iv) (d): 56.4 Sum of new scores = 564 New mean =  \(\frac{564}{10}=56.4\)   (v) (d):   range It is called range of the data.

(i) (a): 14 \(\text { Mean }=\frac{12+8+13+15+22}{5}\) = 14 (ii) (c): 21.1 Since, mean  \((\bar{x})\)  = 14

\( \operatorname{Var}(X) =\frac{1}{n} \sum\left(x_{i}-\bar{X}\right) \) \(=\frac{106}{5}=21.2\)   (iii) (b):   \(\text { Variance }=\sqrt{S . D}\)   (iv) (a): 4.604 \( \mathrm{S} . \mathrm{D} =\sqrt{\mathrm{Var}} \) \(=\sqrt{21.2}=4.604 \)   (v) (c): 10.5 Sum of first 20 natural numbers = S 20 \(=\frac{20}{2}[2 \times 1+(19) 1]\)   s 20 = 210 \(\text { Mean }=\frac{210}{20}\) = 10.5

(i) (a): 0

Therefore,  \(\Sigma d\)   = 0 (ii) (c): 166 \(\Sigma d^{2}\)   = 25 + 1 + 9 + 1 + 36 + 16 + 1 + 9 + 4 + 64 = 166 (iii) (d): 100 n 2 = 10 2 = 100 (iii) (d): –0.006 Rank correlation is given by: \(r_{s}=1-\frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}\)   =  \(1-\frac{6 \times 166}{10\left(10^{2}-1\right)}\)   =  \(1-\frac{996}{990}\)   =  \(\frac{990-996}{990}\)   =  \(\frac{-6}{990}\)   =  –0.006 (v) (b): there is hardly any agreement between the ranks. The very low value (almost 0) indicates that there is hardly any agreement between the ranks.

(i) (b): product of the standard deviations (ii) (c): 80

(iii) (c): 55

(iv) (a): 220

(v) (b): –0.5

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Chapter 13 Class 11 Statistics

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Updated for NCERT 2023-24 Edition.

Get solutions of Chapter 13 Class 11 Statistics of free at teachoo. Answers to all NCERT questions and examples are solved in a step-by-step basis, with detailed explanations of each and every question.

In previous classes, we learned what data is, what mean, median, mode is. 

Mean, median, mode are called central tendency. In this chapter, we will find dispersion. Dispersion means how the data is different from the central tendency. Like if we find mean, how the data is different from the mean.

Like in the figure

Both red and blue graphs have the same mean - 100.

We see that values of the blue graph are very left and right from the mean, that is, they are dispersed from the mean.

And, for the red graph, they are very close to the mean, they are less dispersed from the mean.

This is called Dispersion.

In this chapter, we will study how do we find dispersion. 

The topics in the chapter include

• How to find range
• Ungrouped data
• Grouped data - Discrete frequency deviation and Continuous Frequency Deviation
• And, we will also find Mean deviation about Median for ungrouped and grouped data
• (In Mean Deviation about Mean and Median, we will use the concepts of finding Mean and Median from Class 10 Statistics )
• Then, we study another form of dispersion, known as Standard Deviation and Variance .
• We will learn formulas of Standard Deviation for Discrete Frequency Distribution, Continuous Frequency Distribution and shortcut method as well.
• Then, to compare two groups of numbers, we find Coefficient of Variation.
• And we also do some mix questions like Multiplication of observations, Finding remaining observations, Correcting incorrect observations.

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Unit 13: Statistics

Measures of central tendency: mean, median and mode.

• Statistics intro: Mean, median, & mode (Opens a modal)
• Comparing means of distributions (Opens a modal)
• Choosing the "best" measure of center (Opens a modal)
• Mean, median, and mode Get 5 of 7 questions to level up!

Measures of spread: range, variance, and standard deviation

• Measures of spread: range, variance & standard deviation (Opens a modal)
• Variance of a population (Opens a modal)
• Population standard deviation (Opens a modal)
• The idea of spread and standard deviation (Opens a modal)
• Calculating standard deviation step by step (Opens a modal)
• Mean and standard deviation versus median and IQR (Opens a modal)
• Concept check: Standard deviation (Opens a modal)
• Standard deviation of a population Get 3 of 4 questions to level up!

Other measures of spread: mean absolute deviation

• Range and mid-range (Opens a modal)
• Mean absolute deviation (MAD) (Opens a modal)
• Mean absolute deviation example (Opens a modal)
• Mean absolute deviation (MAD) review (Opens a modal)
• Mean absolute deviation (MAD) Get 3 of 4 questions to level up!

CBSE Class 11 Maths – Chapter 15 Statistics- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Statistics : Notes and Study Materials -pdf

• Concepts of  Statistics
• Statistics Master File
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• NCERT Solution  Statistics
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• Statistics : Solved Example 1

CBSE Class 11 Maths Notes Chapter 15 Statistics

Measure of Dispersion The dispersion is the measure of variations in the values of the variable. It measures the degree of scatteredness of the observation in a distribution around the central value.

Range The measure of dispersion which is easiest to understand and easiest to calculate is the range. Range is defined as the difference between two extreme observation of the distribution. Range of distribution = Largest observation – Smallest observation.

Note: The distribution having a greater coefficient of variation has more variability around the central value, then the distribution having a smaller value of the coefficient 0f variation.

Statistics Class 11 MCQs Questions with Answers

Question 1. The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is (a) 1/5 (b) 2/5 (c) 3/5 (d) 4/5

Answer: (c) 3/5 Hint: Given, ∑x = 12 and ∑x² = 18 Now, varience = ∑x²/n – (∑x/n)² ⇒ varience = 18/10 – (12/10)² ⇒ varience = 9/5 – (6/5)² ⇒ varience = 9/5 – 36/25 ⇒ varience = (9 × 5 – 36)/25 ⇒ varience = (45 – 36)/25 ⇒ varience = 9/25 ⇒ Standard deviation = √(9/25) ⇒ Standard deviation = 3/5

Question 2. The algebraic sum of the deviation of 20 observations measured from 30 is 2. So, the mean of observations is (a) 30.0 (b) 30.1 (c) 30.2 (d) 30.3

Answer: (b) 30.1 Hint: Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2 ⇒ ∑(x i – 30) = 2 {1 ≤ i ≤ 20} ⇒ ∑x i – 30 × 20 = 2 ⇒ (∑x i )/20 – (30 × 20)/20 = 2/20 ⇒ (∑x i )/20 – 30 = 0.1 ⇒ Mean – 30 = 0.1 ⇒ Mean = 30 + 0.1 ⇒ Mean = 30.1

Question 3. The coefficient of variation is computed by (a) S.D/.Mean×100 (b) S.D./Mean (c) Mean./S.D×100 (d) Mean/S.D.

Answer: (b) S.D./Mean Hint: The coefficient of variation = S.D./Mean

Question 4. When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is (a) 1445 (b) 1446 (c) 1447 (d) 1448

Answer: (b) 1446 Hint: Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5 = 7230/5 = 1446

Question 5. If mode of a series exceeds its mean by 12, then mode exceeds the median by (a) 4 (b) 8 (c) 6 (d) 12

Answer: (b) 8 Hint: Given, Mode = Mean + 12 ⇒ Mode – 12 = Mean Now, Mode = 3×Median – 2×Mean ⇒ Mode = 3×Median – 2(Mode – 12) ⇒ Mode = 3×Median – 2×Mode + 24 ⇒ Mode + 2×Mode = 3×Median + 24 ⇒ 3×Mode = 3×Median = 24 ⇒ Mode = Median + 8 So, mode exceeds the median by 8

Question 6. The median and SD of a distributed are 20 and 4 respectively. If each item is increased by 2, the new median and SD are (a) 20, 4 (b) 22, 6 (c) 22, 4 (d) 20, 6

Answer: (c) 22, 4 Hint: Since each value is increased by 2, therefore the median value is also increased by 2. So, new median = 22 Again, the variance is independent of the change of origin. So it remains the same.

Question 7. Range of the data 4, 7, 8, 9, 10, 12, 13 and 17 is (a) 4 (b) 17 (c) 13 (d) 21

Answer: (c) 13 Hint: Give, data are: 4, 7, 8, 9, 10, 12, 13 and 17 Range = Maximum value – Minimum Value = 17 – 4 = 13

Question 8. If Mean = Median = Mode, then it is (a) Symmetric distribution (b) Asymmetric distribution (c) Both symmetric and asymmetric distribution (d) None of these

Answer: (a) Symmetric distribution Hint: In a symmetric distribution, Mean = Median = Mode

Question 9. If the difference of mode and median of a data is 24, then the difference of median and mean is (a) 12 (b) 24 (c) 8 (d) 36

Answer: (a) 12 Hint: Given the difference of mode and median of a data is 24 ⇒ Mode – Median = 24 ⇒ Mode = Median + 24 Now, Mode = 3×Median – 2×Mean ⇒ Median + 24 = 3×Median – 2×Mean ⇒ 24 = 3×Median – 2×Mean – Median ⇒ 24 = 2×Median – 2×Mean ⇒ Median – Mean = 24/2 ⇒ Median – Mean = 12

Question 10. If r is the correlation coefficient, then (a) |r| ≤ 1 (b) r ≤ 1 (c) |r| ≥ 1 (d) r ≥ 1

Answer: (a) |r| ≤ 1 Hint: If r is the correlation coefficient, then |r| ≤ 1

Question 11. If the varience of the data is 121 then the standard deviation of the data is (a) 121 (b) 11 (c) 12 (d) 21

Answer: (b) 11 Hint: Given, varience of the data = 121 Now, the standard deviation of the data = √(121) = 11

Question 12. If the mean of the following data is 20.6, then the value of p is x = 10 15 p 25 35 f = 3 10 25 7 5 (a) 30 (b) 20 (c) 25 (d) 10

Answer: (b) 20 Hint: Mean = ∑ f i × x i /∑ f i ⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5) ⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50 ⇒ 20.6 = (530 + 25p)/50 ⇒ 530 + 25p = 20.6 × 50 ⇒ 530 + 25p = 1030 ⇒ 25p = 1030 – 530 ⇒ 25p = 500 ⇒ p = 500/25 ⇒ p = 20 So, the value of p is 20

Question 13. If the mean of first n natural numbers is 5n/9, then n = (a) 5 (b) 4 (c) 9 (d) 10

Answer: (c) 9 Hint: Given mean of first n natural number is 5n/9 ⇒ (n+1)/2 = 5n/9 ⇒ n + 1 = (5n×2)/9 ⇒ n + 1 = 10n/9 ⇒ 9(n + 1) = 10n ⇒ 9n + 9 = 10n ⇒ 10n – 9n = 9 ⇒ n = 9

Question 14. If one of the observation is zero then geometric mean is (a) (Sum of observation)/n (b) (Multiplication of all observations) n (c) (Multiplication of all observations) 1/n (d) 0

Answer: (d) 0 Hint: Let the observations are 0, a, b, c, ……… up to n Now, geometric mean = (0 × a × b × c × ……… up to n) 1/n = 0 So, geometric mean is 0

Question 15. Which one is measure of dispersion method (a) Renge (b) Quartile deviation (c) Mean deviation (d) all of the above

Answer: (d) all of the above Hint: Range, Quartile deviation, Mean deviation all are the measure of dispersions method.

Question 16. If a variable takes discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5 (x > 0), then the median is (a) x – 5/4 (b) x – 1/2 (c) x – 2 (d) x + 5/4

Answer: (a) x – 5/4 Hint: Given, discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5 Now, arrange them in ascending order, we get x – 7/2, x – 3, x – 5/2, x – 2, x – 1/2, x + 1/2, x + 4, x + 5 Total observations = 8 Now, median = AM of 4th and 5th observations = AM of (x – 2) and (x – 1/2) observations = (x – 2 + x – 1/2)/2 = (2x – 5/2)/2 = x – 5/4

Question 17. If covariance between two variables is 0, then the correlation coefficient between them is (a) nothing can be said (b) 0 (c) positive (d) negative

Answer: (b) 0 Hint: The relationship between the correlation coefficient and covariance for two variables as shown below: r (x, y) = COV(x, y)/{s x × s y } r (x, y) = correlation of the variables x and y COV(x, y) = covariance of the variables x and y s x = sample standard deviation of the random variable x s x = sample standard deviation of the random variable y Now given COV(x, y) = 0 Then r (x, y) = 0

Question 18. The mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is (a) 18 (b) 20 (c) 22 (d) 24

Answer: (b) 20 Hint: Given mean of 100 observations is 20 Now ∑ xi/100 = 20 (1 = i = 100) ⇒ ∑xi = 100×20 ⇒ ∑xi = 2000 3 observations 21, 21 and 18 are recorded in-correctly. So ∑xi = 2000 – 21 – 21 – 18 ⇒ ∑xi = 2000 – 60 ⇒ ∑xi = 1940 Now new mean is ∑ xi/100 = 1940/97 = 20 So, the new mean is 20

Question 19. Varience is independent of change of (a) origin only (b) scale only (c) origin and scale both (d) None of these

Answer: (a) origin only Hint: Varience is independent of change of origin only.

Question 20. Let x 1 , x 2 , x 3 , ……… , x n , be n observations and X be the arithmetic mean. Then formula for the standard deviation is given by (a) ∑(x i – mean)² (b) ∑(x i – mean)2 /n (c) √{∑(x i – mean)²/n} (d) None of these

Answer: (c) √{∑(x i – mean)²/n} Hint: Given, x 1 , x 2 , x 3 , ………. , x n be n observations and X be the arithmetic mean. Now standard deviation = √{∑(x i – mean)²/n}

(Education-Your Door To The Future)

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

In this Post I Have Provided Mathematics  Class 11 Chapter-Wise MCQs, Case-Based Question And Assertion/Reason For Term-1 Session 2021-22 With Solutions. CBSE has Recently Included These Types Of MCQs  And Assertion/Reason For Term-1 Exam 2020.   Every Student Knows these types of Questions Are Very Important For Their Term-1 Examination.

Any student who want to download these MCQs, then they have to click on given respective download links and it will be automatically download in your Google Drive.

CBSE Class 11th  Physics  Term-1 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

For Downlod   Click Here

Given Below Are The Class 11 Mathematics Chapter Name with Respective Download Links Containing Study Material:

CBSE Sample Papers For Class 11 Mathematics

CBSE sample paper for class  11th Maths  is an important tools to analyse itself. Before going to final examinations every student try to solve different types of sample question paper. It will enhance your knowledge and also provide to increase mental ability. Sample question paper are the best resources for the student to prepare for their Board examinations.

These sample question papers are very helpful to the student to get an entire experience before attempting the Board examinations papers. If students solve different types of sample question papers then they get the good confidence about the appropriate answer and they make good score in their Board examinations.

As every student know CBSE always change their exam pattern and also the model paper for their Board examinations that is why it is very important to all the student to understand the Model paper pattern before going to the final examinations.

In this articles I have provided different types of question papers on the basis of CBSE latest exam pattern which will definitely help to the student for the good preparation for their examinations.

In every year before the Board examination, CBSE provide model test paper to the student, on the basis of these model test papers I have prepared sample question papers for class  11th Maths  which will help to the student to understand the concept of model test paper and also it will help to the student for good preparation for their board examination.

CBSE Sample Paper:

Maths And Physics With Pandey Sir  is a website which provide free study material and notes to every students who are preparing for their Board examinations. By going in Level Section you can select CBSE sample papers. In this section, I have provided CBSE sample paper for class 9th to 12th.

Also if you are preparing for any competitive exams, they can find out study material and important notes in Level Section. In this website I have also provided most important MCQs questions, Assertion/Reason based questions and case study based questions for Board examinations or any other competitive examinations.

Benefits Of Solving CBSE Sample Papers for Examination:

Time Management:-

With the help of CBSE sample question papers you can make a proper strategy for their examinations also by practicing sample question papers you can manage your time during the Board examinations.

Examination Strategy:-

After practicing different types of sample question papers you can understand on which section you have spend more time and which section is easy and which section is hard. In this way you can make a proper examination strategy.

Self Evaluation:-

Self evaluation is very important to every student because with the help of self evaluation, student can understand their weak and strong point, which is very important for any Board examinations. After practicing more and more simple paper you can evaluate himself.

To Identify Silly Mistakes:-

As every student know there are many questions in their question paper which are very easy  but student make mistakes in this types of question. By solving different types of sample question papers you can reduce their silly mistakes.

Advantages Of CBSE Sample Question Papers:

* To get a good exam ideas about your examination paper patterns

* Student can easily access to the question papers

* It is very helpful to know about the mark distribution system.

* Helpful in self evaluation

* To know about the pattern of CBSE

* Help to adjust a proper time management

* Available in PDF format

* Chapter wise CBSE sample papers

* CBSE sample papers for class 10

* CBSE sample paper class 12

* CBSE sample paper 2021-22 class 12 Solutions

* CBSE sample paper class 11

* CBSE Sample paper class 9

Important Questions for CBSE Class 11 Maths Chapter 15 – Statistics

Important questions for cbse class 11 maths chapter 15 - statistics, cbse class 11 maths chapter-15 important questions - free pdf download.

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 15 - Statistics prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

1 Marks Questions

1. In a test with a maximum marks 25, eleven students scored 3,9,5,3,12,10,17,4,7,19,21 marks respectively. Calculate the range.

Ans. The marks can be arranged in ascending order as 3,3,4,5,7,9,10,12,17,19,21.

Range = maximum value – minimum value

2. Coefficient of variation of two distributions is 70 and 75, and their standard deviations are 28 and 27 respectively what are their arithmetic mean?

Ans. Given C.V (first distribution) = 70

Similarly for second distribution

3. Write the formula for mean deviation.

4. Write the formula for variance

5. Find the median for the following data.

6. Write the formula of mean deviation about the median

7. Find the rang of the following series 6,7,10,12,13,4,8,12

Ans. Range = maximum value – minimum value

8. Find the mean of the following data 3,6,11,12,18

9. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i 23

11. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans .3x = 6

2x – y = - 3

2 × 2 – y = - 3

- y = - 3 – 4

12. Find the value of 1+i 2 + i 4 + i 6 + i 8 + ---- + i 20

13. Multiply 3-2i by its conjugate.

Ans. Let z = 3 – 2i

14. Find the multiplicative inverse 4 – 3i.

Ans. Let z = 4 – 3i

17. If 1, w, w 2 are three cube root of unity, show that (1 – w + w 2 ) (1 + w – w 2 ) = 4

Ans. (1 – w + w 2 ) (1 + w – w 2 )

(1 + w 2 - w) (1 + w – w 2 )

19. Write the real and imaginary part 1 – 2i 2

Ans. Let z = 1 – 2i 2

=1 – 2 (-1)

Re (z) = 3, Im (z) = 0

20. If two complex number z 1 , z 2 are such that |z 1 | = |z 2 |, is it then necessary that z 1 = z 2

Ans. Let z 1 = a + ib

22. Find the number of non zero integral solution of the equation |1-i| x = 2 x

Which is false no value of x satisfies.

23. If (a + ib) (c + id) (e + if) (g + ih) = A + iB then show that

4 Marks Questions

1.The mean of 2,7,4,6,8 and p is 7. Find the mean deviation about the median of these observations.

The near of these observations is 7

Arrange the observations in ascending order 2,4,6,7,8,15

Calculation of mean deviation about Median.

2.Find the mean deviation about the mean for the following data!

Mean deviation about the mean

Ans. The given numbers are 1, 2, 3, ……, n

4.Find the mean variance and standard deviation for following data

Note: - 4 th , 5 th and 6 th columns are filled in after calculating the mean.

5.The mean and standard deviation of 6 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.

As each observation is multiplied by 3, new observations are

6.Prove that the standard deviation is independent of any change of origin, but is dependent on the change of scale.

7.Calculate the mean deviation about the mean for the following data

Expenditure 0-100100-200200-300300-400400-500500-600600-700700-800

persons 489107543

8.Find the mean deviation about the median for the following data

Marks 0-1010-2020-3030-4040-5050-60

No. of boys 810101642

9.An analysis of monthly wages point to workers in two firms A and B, belonging to the same industry, given the following result. Find mean deviation about median.

Firm AFirm B

No of wages earns586648

Average monthly wagesRs 5253Rs 5253

Ans. For firm A, number of workers = 586

Average monthly wage is Rs 5253

= Rs 3078258

=Rs 3403944

Hence firm B pays out amount of monthly wages.

10.Find the mean deviation about the median of the following frequency distribution

Class 0-66-1212-1818-2424-30

Frequency8101295

12-18 is the medias class

11.Calculate the mean deviation from the median from the following data

Salary per week(in Rs) 10-2020-3030-4040-5050-6060-70

no. of workers 461020106

40-50 is the median class

Taking conjugate both side

Ans. i 25 + (1+3i) 3

20.For what real value of x and y are numbers equal (1+i) y 2 + (6+i) and (2+i) x

Ans. (1+i) y 2 + (6 + i) = (2 + i) x

y 2 + iy 2 + 6 + i = 2x + xi

(y 2 + 6) + (y 2 + 1) i = 2x + xi

y 2 + 6 = 2x

y 2 + 1 = x

y 2 = x – 1

x – 1 + 6 = 2x

taking conjugate both side

x 2 + y 2 = 1

23.Find the real values of x and y if (x - iy) (3 + 5i) is the conjugate of – 6 – 24i

(x – iy) (3 + 5i) = - 6 + 24i

3x + 5xi – 3yi – 5yi 2 = - 6 + 24i

6 Marks Questions

1.Calculate the mean, variance and standard deviation of the following data:

2.The mean and the standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who mistook one observation as 50 instead of 40. What are the correct mean and standard deviation?

= incorrect sum of observation =4000

= correct sum of observations = 4000-50+40

Using incorrect values,

= 162601-2500+1600=161701

Hence, correct mean is 39.9 and correct standard deviation is 5.

3.200 candidates the mean and standard deviation was found to be 10 and 15 respectively. After that if was found that the scale 43 was misread as 34. Find the correct mean and correct S.D

= 8000-34+43= 8009

4.Find the mean deviation from the mean 6,7,10,12,13,4,8,20

5.Find two numbers such that their sum is 6 and the product is 14.

Ans. Let x and y be the no.

Class 11 Maths Chapter 15 Statistics MCQs

Class 11 Maths Chapter 15 Statistics MCQs with solutions are provided at BYJU’S for students. The multiple-choice questions for Class 11 Maths Chapter 15 Statistics are prepared as per the latest exam pattern by our subject experts. The objective type questions are given here, as per the CBSE syllabus (2022 – 2023) and NCERT curriculum. By solving these MCQs on Statistics, students can have a revision for the chapter and also score better in the exam.

Also check: Class 11 Maths chapter-wise MCQs

MCQs for Class 11 Maths Chapter 15 Statistics

Class 11 Maths MCQs of Chapter 15 Statistics are provided here with the correct option. These multiple-choice questions are given here with proper explanations. The questions are based on the Statistics concept and related formulas. Students are advised to solve the MCQ questions on Statistics by themselves and then verify with the provided answers.

Download PDF – Chapter 15 Statistics MCQs

————————————————————————————————————————–

Q.1: Range of a data is equal to:

(a) Range = Max Value – Min Value

(b) Range = Max Value + Min Value

(c) Range = (Max Value – Min Value)/2

(d) Range = (Max Value + Min Value)/2

Answer: (a) Range = Max Value – Min Value

Q.2: Relation between mean, median and mode is given by:

(a) Mode = 2 Median – 3 Mean

(b) Mode = 2 Median + 3 Mean

(c) Mode = 3 Median – 2 Mean

(d) Mode = 3 Median + 2 Mean

Answer: (c) Mode = 3 Median – 2 Mean

Q.3: If the variance of the data is 121, the standard deviation of the data is:

Answer: (b) 11

Explanation:

Given, variance of the data = 121

Now, the standard deviation of the data = √(121)

Q.4: The geometric mean of series having mean = 25 and harmonic mean = 16 is:

Answer: (b) 20

The relationship between Arithmetic Mean (AM), Geometric Mean (GM) And Harmonic Mean (HM) is

GM² = AM × HM

Given AM = 25

So GM² = 25 × 16

⇒GM = √(25 × 16)

So, Geometric Mean = 20

Q.5: The coefficient of correlation r satisfies:

(a) I r I ≤ 1

(b) 0 < r < 1

(c) I r I > 1

(d) − 1 < r < 0

Answer: (a) I r I ≤ 1

Q.6: The coefficient of variation is computed by:

(i) S.D/.Mean×100

(ii) S.D./Mean

(iii) Mean./S.D×100

(iv) Mean/S.D.

Answer: (ii) S.D./Mean

Q.7: Find the median of 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Answer: C. 47.5

Explanation: First we have to arrange the given observations into ascending order,

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations is 10

Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2

(10/2)th observation = 5th = 46

(10/2)+ 1)th observation = 5 + 1

Median = (46 + 49)/2

Q.8: Find the mean of 6, 7, 10, 12, 13, 4, 8, 12.

Answer: A. 9

Explanation: Mean = Sum of observations ÷ Number of observations

= (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

Q.9: If the variance is 625, what is the standard deviation?

D. None of the above

Answer: C. 25

Explanation: Given,

Variance, σ 2 = 625

Standard deviation, σ = √625 = 25

Q.10: If the mean of first n natural numbers is 5n/9, n =

Answer: (c) 9

Given, mean of first n natural number is 5n/9

⇒ (n+1)/2 = 5n/9

⇒ n + 1 = (5n×2)/9

⇒ n + 1 = 10n/9

⇒ 9(n + 1) = 10n

⇒ 9n + 9 = 10n

⇒ 10n – 9n = 9

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