Statistics Class 11th Notes - Free NCERT Class 11 Maths Chapter 15 notes - Download PDF

Statistics Class 11th Notes - Free NCERT Class 11 Maths Chapter 15 notes - Download PDF

Edited By Ramraj Saini | Updated on Mar 22, 2022 05:45 PM IST

Class 11 Math chapter 15 notes are regarding statistics. In chapter 15 we will be going through the mean, mode, and median concepts in Statistics Class 11 notes. This Class 11 Maths chapter 15 notes contains the following topics: statistics, range, mean, median, mode, mean deviation about mean and median in grouped and ungrouped data, continuous, discrete distributions, the standard deviation for ungrouped, continuous, discrete distributions, variance, coefficient of variance with standard formulas. NCERT Class 11 Math chapter 15 contains a detailed explanation of topics like - mean, median, mode, frequency, variation. Important topics are explained in CBSE Class 11 Maths chapter 15 notes in a very simple and detailed way. All these notes can be downloaded from Class 11 Maths chapter 15 notes pdf download, Statistics Class 11 notes, and Class 11 Statistics notes pdf download.

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  1. NCERT CLASS 11 CHAPTER 15 NOTES
  2. Significance of NCERT Class 11 Maths Chapter 15 Notes:
  3. NCERT Class 11 Notes Chapter Wise.

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NCERT CLASS 11 CHAPTER 15 NOTES

Statistics:

Collecting and analysing data in different forms is called statistics.

Measure of dispersion:

Measuring the difference or the variations in the values is called dispersion measurement.

Range:

The difference between the highest and the lowest values gives us the range.

Mean:

Mean is generally the ratio between the sum of observations to the number of observations.

Denoted by x.

Mathematical representation: \frac{1}{n} \sum_{i=1}^nx_i

n denotes the number of observations.

xi denotes i th observations.

Median:

The middle most term in all the observations after arranging them in either ascending or descending order denotes the Median.

For the odd number of observations :

Median = {n+1}/2 th observation

For an even number of observations:

It is the mean of n/2 th and {n+1}/2 th observation.

Mode:

The observation that repeats most frequently is the mode of the data.

Mean Deviation for ungrouped data:

Let n be the observation: x1,x2,x3,............,xn

Mean Deviation about Mean:

MD ( \bar{x} ) = \frac{ \sum |x_i-x|}{n}

Here x is the mean.

Steps to calculate mean deviation about a mean:

Step 1: Calculate the means value x‾ the formula.

\frac{1}{n} \sum_{i=1}^nx_i

Step 2: Find the value of deviation from xi to x.

Step 3: Find the value of mean of absolute value from the formula specified above under the definition.

Mean deviation about Median:

MD (M) = \sum \frac {|x_i-M|}{n}

Here M is the median.

Mean Deviation for grouped data:

It can be of two types:

  1. Discrete distribution

  2. Continuous distribution

Mean Deviation about mean for discrete distribution:

Let n be the observation: x1, x2, x3,............,xn, and the frequencies are

f1,f2,f3,.............,fn respectively.

MD (\bar{x}) = \frac{ \sum f_i |x_i-\bar{x}|}{\sum f_i} = \frac{\sum f_i |x_i-\bar{x}|}{N}

Here, xi and fi are respective observations and frequencies.

N is the sum of frequencies.

Mean Deviation about median for discrete distribution:

MD (M) = \frac{ \sum f_i |x_i-M|}{\sum f_i} = MD (\bar{x}) = \frac{ \sum f_i |x_i-M|}{N}

Here, M denotes the median.

Mean Deviation about mean for continuous distribution:

Let n be the observation: x1,x2,x3,............,xn, and the frequencies are

f1,f2,f3,.............,fn respectively.

MD (\bar{x}) = \frac{ \sum f_i |x_i-\bar{x}|}{\sum f_i} = MD (\bar{x}) = \frac{ \sum f_i |x_i-\bar{x}|}{N}

Step\ deviation : x = a+ \frac{\sum_{i=1}^n f_id_i}{N} h

Where we have di = {xi-a}/h

Here, a is the assumed mean.

h is the interval between the upper and lower class frequencies.

Mean Deviation about median for continuous distribution:

MD (M) = \frac{ \sum { f_i |x_i-M}|}{ \sum f_i} = MD (\bar{x}) = \frac{ \sum { f_i |x_i-M|}}{N}

Here, M denotes median.

M= l+ \frac{\frac{N}{2}- C}{f} h

Here, l is the lowest boundary value

N is the sum of frequencies

C is the value of cumulative frequency of selected class

f is selected frequency

h is the interval between the upper and lower class frequencies.

To overcome a few limitations of the mean deviation we use the concepts called variance and standard deviation.

Variance:

It is the square of Standard deviation.

\sigma^2=\frac{1}{n} \sum |x_i -\bar{x}|^2

σ2= variance

x¯ = mean

n = sum of frequencies

Standard Deviation:

It is the square root of variance.

\sigma=\sqrt{\frac{1}{n} \sum |x_i -\bar{x}|^2}

Standard deviation of discrete frequencies:

\sigma=\sqrt{\frac{1}{N} \sum f_i |x_i -\bar{x}|^2}

Standard deviation of continuous frequencies:

\sigma=\frac{1}{N} \sqrt{ \sum N f_i {x_i}^2 - \sum (f_ix_i)^2}

Another formula:

\sigma=\frac{h}{N} \sqrt{ \sum N f_i {d_i}^2 - \sum (f_id_i)^2}

Here,

h is the width of the class

di = (xi-A)/h where A is assumed mean

Coefficient of variance:

\\ CV = \frac{Standard \ deviation} {Mean} 100 \\ \\ CV = \frac{\sigma}{\bar{x}} 100

Note: When CV is coefficient of variance is greater then we have greater variance than that of smaller variance.

With this topic we conclude NCERT Class 11 chapter 15 notes.

The link for NCERT textbook pdf is given below:

URL: ncert.nic.in/textbook/pdf/kemh115.pdf

Significance of NCERT Class 11 Maths Chapter 15 Notes:

NCERT Class 11 Maths chapter 15 notes will be very much helpful for students to score maximum marks in their 11 board exams. In Statistics Class 11 chapter 15 notes we have discussed many topics: statistics, range, mean, median, mode, mean deviation about mean and median in grouped and ungrouped data, continuous, discrete distributions, the standard deviation for ungrouped, continuous, discrete distributions, variance, coefficient of variance with standard formulas. NCERT Class 11 Maths chapter 15 is also very useful to cover major topics of the Class 11 CBSE Maths Syllabus.

The NCERT Class 11 Maths chapter 15 notes are in compact form. The statistics chapter is explained in detail in the notes. Most useful topics and formulas are explained in an easy manner. The students can download pdf from the Statistics Class11 chapter 15 pdf download link.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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