Statistics, Class 10, Maths Detailed Chapter Notes PDF Download

INTRODUCTION

In class IX, we have studied about the presentation of given data in the form of ungrouped as well as grouped frequency distributions. We have also studied how to represent the statistical data in the form of various graphs such as bar graphs, histograms and frequency polygons. In addition, we have studied the measure of central tendencies such as mean, median and mode of ungrouped data.

In this chapter, we shall discuss about mean, median and mode of grouped data. We shall also discuss the
concept of cumulative frequency, cumulative frequency distribution and cumulative frequency curve (ogive).

MEAN OF UNGROUPED DATA

We know that the mean of observations is the sum of the values of all the observations divided by the total
number of observations i.e., If x₁, x₂, x₃,......x_n are n observations, then
mean,  or   denotes the sum x₁, + x₂, + x₃ + ,...... + x_n.

The arithmetic mean of grouped data may also be calculated by any one of the following methods :
1. Direct method

2. Short-cut method or Assumed-mean method

3.Step-deviation method.

MEAN OF GROUPED DATA

DIRECT METHOD

If x₁, x₂, x₃,......x_n are observations with respective frequencies then mean, defined by

 or  

To find mean of grouped Data

The following steps should be followed in finding the arithmetic mean of grouped data by direct method.

STEP - 1 :Find the class mark (xi) of each class using, 

STEP - 2 : Calculate f_ix_i for each i

STEP - 3 : Use the Formula : mean, 

SHORTCUT METHOD OR ASSUMED MEAN METHOD

In this case, to calculate the mean, we follow the following steps :

STEP-1 :Find the class mark x_i of each class using

STEP-2 :Choose a suitable value of x_i in the middle as the assumed mean and denote it by 'a'.

STEP-3 :Find d_i = x_i – a for each i

STEP-4 :Find f_i × d_i for each i

STEP-5 :Find   fi

STEP-6 :Calculate the mean, (x) by using the formula 

STEP-DEVIATION METHOD

Sometimes, the values of x and f are so large that the calculation of mean by assumed mean method becomes quite inconvenient. In this case, we follow the following steps:

STEP-1 :Find the class mark xi of each class by using

STEP-2 :Choose a suitable values of xi in the middle as the assumed mean and denote it by 'a'.

STEP-3 :Find h = (upper limit –lower limit) for each class.

STEP-4 :Find  for each class.

STEP-5 :Find f_iu_i for each i.

STEP-6 :Calculate, the mean by using the formula

Ex.1 Find the mean of the following data :

Sol. We may prepare the table as given below :

Ex.2 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Sol. We may prepare the table as given below :

Given, mean = 18

  ⇒ 752 + 18f = 752 + 20f  ⇒ 2f = 40 ⇒f = 20

2f = 40 ⇒ f = 20

Ex.3 Find the missing frequencies f₁ and f₂ in the table given below, it is being given that the mean of the given frequency distribution is 50.

Sol. We may prepare the table as given below :

Given, mean = 50

⇒ 3400 + 50f₁+ 50f₂ = 3480 + 30f₁ + 70f₂

⇒   20f₁ - 20f₂ = 80 ⇒ f₁ - f₂ = 4   .......(i)
And ∑f_i = 68 + f₁+ f₂
∴ 120 = 68 + f₁ + f₂    [∵ ∑f₁ = 120]
⇒ f₁ + f₂ = 52    ...(ii)
Adding (1) and (2), we get 2f₁ = 56 ⇒ f₁ =28

∴ f₂ = 24

Hence, the missing frequencies f₁ and f₂ are 28 and 24 respectively.

Ex.4 The following table gives the marks scored by 100 students in a class test :

Find the mean marks scored by a student in class test.

Sol. We may prepare the table with assumed mean, a = 35 as given below :

Ex.5 Thirty women were examined in a hospital by a doctor and the number of heart beats per minute, were recorded and summarised as follows. Find the mean heart beats per minute for these women, by using assumed mean method.

Sol. We may prepare the table with assumed mean, a = 75.5 as given below :

Ex.6 Find the mean of the following distribution by step-deviation method :

Sol. We may prepare the table with assumed mean a = 120 and h = 20 as given below :

Class	Frequency (fi)	Class inailc (xi)		fiui
50-70	18	60	-3	-54
70-90	12	80	_2	-24
90-110	13	100	-1	-13
110-130	27	120 = a	0	0
130-150	8	140	1	8
150-170	22	160	2	44
	N= 100			∑fiui = -39

Ex.7 Find the mean marks from the following data :

Sol. We may prepare the table as given below :

Ex.8 Find the mean marks of students from the adjoining frequency distribution table.

Sol. We may prepare the table as given below :

   = 51.125 = 51.1 (approx)

Ex.9 Find the arithmetic mean of the following frequency distribution.

Sol. The given series is in inclusive form. We may prepare the table in exclusive form with assumed mean a = 42 as given below :

MEDIAN OF A GROUPED DATA

MEDIAN : It is a measure of central tendency which gives the value of the middle most observation in the data. In a grouped data, it is not possible to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves.

MEDIAN CLASS : The class whose cumulative frequency is greater than N/2 is called the median class.

To calculate the median of a grouped data, we follow the following steps :

STEP-1 :Prepare the cumulative frequency table corresponding to the given frequency distribution and obtain
N = ∑f_i.

STEP-2 :Find N/2

STEP-3 :Look at the cumulative frequency just greater than N/2 and find the corresponding class (Median class).

STEP-4 :Use the formula Median,

Where ℓ = Lower limit of median class.
f = Frequency of the median class.
C = Cumulative frequency of the class preceding the median class.
h = Size of the median class.

 Ex.10 Find the median of the following frequency distribution :

Sol. At first we prepare a cumulative frequency distribution table as given below :

Her , N = 100

∴ N/2 = 50

The cumulative frequency just greater than 50 is 64 and the corresponding class is 20-30.

So, the median class is 20-30.

∴ ℓ = 20, N = 100, C = 28, f = 36 and h = 10

Therefore, median