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

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Statistics Class 11 Notes

Careers360 experts have curated these Statistics Class 11 Notes to help students revise quickly and confidently.

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: $\frac{1}{n} \sum_{i=1}^n x_i$
n denotes the number of observations.
$x_i$ denotes $i$th observations.

Median

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

For the odd number of observations :

Median $=\left(\frac{n+1}{2}\right)$th observation
For an even number of observations:
It is the mean of $\left(\frac{n}{2}\right)$ th and $\left(\frac{n+1}{2}\right)$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: $x_1, x_2, x_3, \ldots \ldots \ldots, x_n$

Mean Deviation about Mean

$
M D(\bar{x})=\frac{\sum\left|x_i-x\right|}{n}
$

Here, x is the mean.
Steps to calculate the mean deviation about a mean:
Step 1: Calculate the mean value $\bar{x}$ using the formula.

$
\frac{1}{n} \sum_{i=1}^n x_i
$

Step 2: Find the value of deviation from $\mathrm{x}_{\mathrm{i}}$ to x.

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

Mean deviation about Median

$
M D(M)=\sum \frac{\left|x_i-M\right|}{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 the mean for a discrete distribution

Let $n$ be the observation: $x_1, x_2, x_3, \ldots \ldots \ldots, x_n$, and the frequencies are $f_1, f_2, f_3,.......,f_n$ respectively.

$
M D(\bar{x})=\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}=\frac{\sum f_i\left|x_i-\bar{x}\right|}{N}
$
Here, $x_i$ and $f_j$ are the respective observations and frequencies.
$N$ is the sum of frequencies.

Mean Deviation about median for a discrete distribution

$
M D(M)=\frac{\sum f_i\left|x_i-M\right|}{\sum f_i}=M D(\bar{x})=\frac{\sum f_i\left|x_i-M\right|}{N}
$
Here, $M$ denotes the median.

Mean Deviation about the mean for a continuous distribution

Let n be the observation: $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots \ldots \ldots \ldots . \mathrm{x}_n$, and the frequencies are $\mathrm{f}_1, \mathrm{f}_2, \mathrm{f}_3, \ldots \ldots \ldots \ldots . . \mathrm{f}_{\mathrm{n}}$ respectively.

$
\begin{gathered}
M D(\bar{x})=\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}=M D(\bar{x})=\frac{\sum f_i\left|x_i-\bar{x}\right|}{N} \\
\end{gathered}
$
$\text { Step deviation}: x=a+\frac{\sum_{i=1}^n f_i d_i}{N} h$
Where we have $d_i=\frac{\left(x_i-A\right)}{h}$
Here, $a$ is the assumed mean, $h$ is the interval between the upper and lower class frequencies

Mean Deviation about median for continuous distribution

$
M D(M)=\frac{\sum f_i\left|x_i-M\right|}{\sum f_i}=M D(\bar{x})=\frac{\sum f_i\left|x_i-M\right|}{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 the cumulative frequency of the selected class
$f$ is the 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 the Standard deviation.

$
\begin{aligned}
& \sigma^2=\frac{1}{n} \sum\left|x_i-\bar{x}\right|^2 \\
& \sigma^2=\text { variance } \\
& \bar{x}=\text { mean } \\
& n=\text { sum of frequencies }
\end{aligned}
$

Standard Deviation

It is the square root of variance.

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

Standard deviation of discrete frequencies

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

Standard deviation of continuous frequencies

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

Another formula:

$
\sigma=\frac{h}{N} \sqrt{\sum N f_i d_i^2-\sum\left(f_i d_i\right)^2}
$
Here, $h$ is the width of the class

$
d_i=\frac{\left(x_i-A\right)}{h} \text {, where } \mathrm{A} \text { is assumed mean }
$

Coefficient of variance

$
\begin{array}{lc}
C V =\frac{\text { Standard deviation }}{\text { Mean }}\times 100 \\
C V =\frac{\sigma}{\bar{x}} \times100
\end{array}
$

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

With this topic, we conclude the NCERT Class 11 Statistics notes.

Statistics: Previous Year Question and Answer

Question 1:

If x_i’s are the midpoints of the class intervals of grouped data, fi’s are the corresponding frequencies and $\bar{x}$ is the mean, then $\Sigma\left(f_{i} x_{i}-\bar{x}\right)$ is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Solution:
Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean $\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{n}$
By cross multiplication, we get,
$\sum f_{i}x_{i}= \bar{x}n\cdots (1)$
$\sum \left ( f_{i}x_{i} -\bar{x}\right )= \sum f_{i}x_{i} -\sum \bar{x}$
$=n\bar{x}-\sum \bar{x}$ (from equation (1))
$=n\bar{x}-n\bar{x}$
= 0

Hence, the correct answer is 0.

Question 2:

The abscissa of the point of intersection of the less-than-type and the more-than-type cumulative frequency curves of a grouped data gives its
(A) mean
(B) median
(C) mode
(D) all three above

Solution:
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarise categorical variables.
If we make a graph of less than type and of more than type grouped data and find the intersection point, then the value at the abscissa is the median of the grouped data.
Hence, option (B) is correct.

Question 3:

For the following distribution:

The sum of the lower limits of the median class and the modal class is
(A) 15
(B) 25
(C) 30
(D) 35

Solution:
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarise categorical variables.

$\frac{N}{2}=\frac{66}{2}=33$ which ies in the class 10-15.
Hence, the median class is 10 – 15.
The class with the maximum frequency is the modal class, which is 15 – 20
The lower limit of the median class = 10
The lower limit of the modal class = 15
Sum = 10 + 15 = 25

Hence, the correct answer is 25.

Importance of NCERT Class 11 Maths Chapter 13 Notes:

NCERT Class 11 Maths Chapter 13 Notes play a vital role in helping students grasp the core concepts of the chapter easily and effectively, so that they can remember these concepts for a long time. Some important points of these notes are:

• Effective Revision: These notes provide a detailed overview of all the important theorems and formulas, so that students can revise the chapter quickly and effectively.
• Clear Concepts: With these well-prepared notes, students can understand the basic concepts effectively. Also, these notes will help the students remember the key concepts by breaking down complex topics into simpler and easier-to-understand points.
• Time Saving: Students can look to save time by going through these notes instead of reading the whole lengthy chapter.
• Exam Ready Preparation: These notes also highlight the relevant contents for various exams, so that students can get the last-minute minute very useful for exams.

NCERT Class 11 Notes Chapter Wise

For students' preparation, Careers360 has gathered all Class 11 Maths NCERT Notes here for quick and convenient access.

NCERT Notes Subject-Wise

Given below are some subject-wise links for the NCERT Notes for class 11.

• NCERT Notes Class 11 Physics
• NCERT Notes Class 11 Chemistry
• NCERT Notes Class 11 Biology

Subject-Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

• NCERT Exemplar Class 11 Solutions
• NCERT Exemplar Class 11 Maths
• NCERT Exemplar Class 11 Physics
• NCERT Exemplar Class 11 Chemistry
• NCERT Exemplar Class 11 Biology

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

• NCERT Solutions for Class 11 Mathematics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Physics

• NCERT Solutions for Class 11 Biology

NCERT Books and Syllabus

Students should always analyse the latest CBSE syllabus before making a study routine. The following links will help them check the syllabus. Also, here is access to more reference books.

• NCERT Book for Class 11
• NCERT Syllabus for Class 11