Statistics Class 10th Notes - Free NCERT Class 10 Maths Chapter 14 Notes - Download PDF

# Statistics Class 10th Notes - Free NCERT Class 10 Maths Chapter 14 Notes - Download PDF

Edited By Ramraj Saini | Updated on Mar 19, 2022 02:07 PM IST

The Class 10 Maths chapter 14 notes give the idea of chapter 14 in the NCERT book. Statistics Class 10 notes is the summary of the chapter and CBSE Class 10 Maths chapter 14 notes discuss the Statistics of diagrams, an important aspect of geometry. The primary topics covered in Class 10 Maths chapter 14 notes are the Direct method, the Assumed mean method, Step-deviation method, Median, and mode. These are the important topics in NCERT Class 10 Maths chapter 14 notes. Examples are also covered in notes for Class 10 Maths chapter 14. These given notes - NCERT notes for Class 10 Maths chapter 14 are very helpful for revision. Statistics notes are important for the CBSE Exam and can be downloaded from Class 10 Maths chapter 14 notes pdf download or Statistics Class 10 notes pdf download.

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## NCERT Class 10 Maths Chapter 14 Notes

Mean of Grouped Data: If n observations in the raw data consist of only k distinct values, denoted by

x1, x2,.....xk of the observed variable x, occurring with frequencies f1, f2……fk respectively. Then, the formula for arithmetic mean is:

So, we can express this formula as

Where

denotes the total frequency. This method is also known as the Direct method.

Example1. Find the mean from the following data:

 Marks No. of Students 20 3 15 2 10 5

Solution:

 Marks No. of Students 20 3 60 15 2 30 10 5 50 Total = 10 =140

• Class interval type question

Example 2. Calculate the arithmetic mean from the following data

 Marks No. of students Less than 10 4 Less than 20 13 Less than 30 18 Less than 40 9 Less than 50 6

Solution:

 Marks No. of students Mid-point 0-10 4 20 10-20 13 195 20-30 18 450 30-40 9 315 40-50 6 270 Total = 50

## Assumed Mean Method or Shortcut Method

In this method, we choose an arbitrary constant a (also called the assumed mean) and subtract it from each of the values xi. The reduced value (xi-a) is called the deviation of xi from a and is denoted by di, where a is somewhere in the middle of all values of xi

Therefore the mean by the short-cut method is given by

Step-Deviation Method for computing the mean

In this method the reduced values (xi-a) i.e., deviations are divided by a constant k, where h is generally taken to be the class interval in the frequency table. Thus, we define

The formula for arithmetic mean is given by

Where

Example 3. Find the arithmetic mean of the following frequency distribution using the Assumed mean method nearest to a whole number.

 Class- Interval Frequency 10-1515-2020-2525-3030-3535-40 5781294

Solution: Let the assumed mean (A) = 27.5

 Class-Interval Frequencyf Mid-value 10-1515-2020-2525-3030-3535-40 5781294 12.517.522.527.5 a.32.537.5 -15-10-50+5+10 -75-70-4004540 Total

Example 4. Find the mean from the following data by using the step deviation method

 Class interval 0-10 10-20 20-30 30-40 40-50 Frequency 6 10 9 4 11

Let the assumed mean(a) = 25 and

The class interval (h) = 10

 Class interval Frequency Mid-point 0-10 6 5 -20 -2 -12 10-20 10 15 -10 -1 -10 20-30 9 25 0 0 0 30-40 4 35 10 1 4 40-50 11 45 20 2 22 Total

Here,

Where

Median from ungrouped Data

The median in a set of numbers is the middle value (or the mean of two middle values) when the numbers are arranged in order of magnitude.

For the ungrouped data, the median is computed as follows:

Step 1. The values of the variate are arranged in order of magnitude.

Step 2. The middle-most value is taken as the median.

If the numbers of terms, n, in the raw data are odd, then the median will be the

the term, when arranged in order of magnitude. In this case, there will be one and only one value of the median.

On the other hand, if the numbers of terms, n, in the raw data are even, then the data arranged in order of magnitude will have two middle-most values (terms), viz. n/2 th and (n/2+1)th terms. In this case, any value between these two middle-most terms can be taken as the median or the arithmetic mean of these two middle terms given the required media, i.e.

Median Of Discrete Series

First, arrange the terms in ascending or descending order. Then, prepare a cumulative frequency table. Let the total frequency be n.

1. If n is odd, then

1. If n is even, then

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Median=12Size of the n2th term + Size of the [n2+1]th term

Example 5. The median of 5, 7, 15, 17, 9, 19, 11 and 17 is:

(i) 9 (ii) 12 (iii) 13 (iv) 17

Ans. Option (iii) 13.

Solution : Arrange the data 5, 7, 15, 17, 9, 19, 11 and 17 in ascending order:

5, 7, 9, 11, 15, 17, 17, 19

Number of terms (N) = 8 (which is even)

Median term=12n2th term+n2+1th term

=124thterm +5thterm

=1211+15

=1226=13

Example 6. The median of 7, 6, 9, 10, 8, 12 and 13 is:

(i) 9 (ii) 10 (iii) 8 (iv) 12.

Ans. (i) 9

Solution: Arrange the numbers 7, 6, 9, 10, 8, 12, and 13 in ascending order, and we get

6, 7, 8, 9, 10, 12, 13

Here, the number of terms (N) = 7, (which is odd)

Therefore,

• Cumulative frequency: It is the running total of all frequencies.

Median from Grouped Data

If the data is grouped, first make the cumulative frequency table. Locate n/2th item. The class in which this frequency occurs is called the median class interval. Then put

L₁ = lower limit of the median class-interval

h = class size

f = frequency of the median class-interval

n = total frequency

c.f = cumulative frequency of the class interval preceding the median class.

Apply the formula :

It is to be noted that when the data is grouped and the series is continuous, the median is the size of

n/2th item and not (n+1)/2th item (if the sum i.e., ∑f = n = even).

Example 6. Find the median for the following frequency table:

 Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 4 13 18 9 6

Solution :

 Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 4 13 18 9 6 Cumulative Frequency 4 13 + 4 =17 18 + 17 =35 35 + 9 = 44 44 + 6 = 50

With comes in the class interval 20 - 30

Therefore, median class = 20 - 30

L₁ = 20, h = 10, f = 18, n/2 = 25, c.f = 17

Example 7. Find the median for the following frequency distribution:

 Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 6 9 12 8 15

Solution :

 Class Interval Frequency Cumulative Frequency 0 - 10 6 6 10 - 20 9 15 20 - 30 12 27 30 - 40 8 35 40 - 50 15 50 Total N = 50

With comes in the class interval 20 - 30

Therefore, median class = 20 - 30

By the formula,

Mode: Mode is the value of that item in the series which occurs most predominantly, i.e., comes a maximum number of times.

In ungrouped data:- It is that value of a variable that has the maximum frequency.

It finds application in practical life; for e.g., in business. A factory produces more shoes of that size which is worn by the maximum number of people, i.e., which is the modal size.

In grouped data:- In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:

f1= frequency of Modal class

f0= frequency of the class preceding the modal class

f2= frequency of the class succeeding the modal class.

l= lower limit of the modal class

h= size of the class interval (assuming all class sizes to be equal)

Example 8.Find the mode of the numbers 23,14,10,12,11,12,23,20,18,12,10,12, and 23.

Solution: The frequency of 12 is the maximum among the given numbers.

Mode = 12

Example 9. : A group of students took a survey of 20 families in a neighborhood, that yielded the following frequency table for the number of family members in a household:

 Family size 1 - 3 3 - 5 5 - 7 7 - 9 9 - 11 Number of families 7 8 2 2 1

Find the mode of this data.

Solution: Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 – 5.

Now

modal class = 3 – 5,

lower limit (l) of modal class = 3, class size

(h) = 2

Frequency ( f 1 ) of the modal class = 8,

Frequency ( f 0 ) of class preceding the modal class = 7,

Frequency ( f 2 ) of class succeeding the modal class = 2.

Now, let us substitute these values in the formula:

Relation between Mean, Median, and Mode

## Significance of NCERT Class 10 Maths Chapter 14 Notes

Statistics Class 10 notes will help to understand the formulas, statements, rules in detail. Also, these NCERT Class 10 Maths chapter 14 notes contain previous year’s questions and NCERT TextBook pdf. Class 10 Maths chapter 14 notes pdf download or Statistics Class 10 notes pdf download is helpful for offline preparation.

## Class 10 Chapter Wise Notes

 NCERT Class 10 Maths Chapter 1 Notes NCERT Class 10 Maths Chapter 2 Notes NCERT Class 10 Maths Chapter 3 Notes NCERT Class 10 Maths Chapter 4 Notes NCERT Class 10 Maths Chapter 5 Notes NCERT Class 10 Maths Chapter 6 Notes NCERT Class 10 Maths Chapter 7 Notes NCERT Class 10 Maths Chapter 8 Notes NCERT Class 10 Maths Chapter 9 Notes NCERT Class 10 Maths Chapter 10 Notes NCERT Class 10 Maths Chapter 11 Notes NCERT Class 10 Maths Chapter 12 Notes NCERT Class 10 Maths Chapter 13 Notes NCERT Class 10 Maths Chapter 14 Notes NCERT Class 10 Maths Chapter 15 Notes

## NCERT Class 10 Exemplar Solutions for Other Subjects:

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