Measure of Central Tendency for Grouped Data

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Mean/Average

It is defined as the ratio of sum of all observations to the number of observations.

Mean	=	Sum of all observations
		Number of observations

For an ungrouped frequency distribution, the Mean,

͞x	=	fi xi
		fi

where,
f_i × x_i = Sum of all observations
f_i = Total number of Observations

↪ In most of real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data.
↪ In Grouped frequency distribution, observations are classified into class intervals of same widths.
↪ By convention, the common observation belongs to the higher class, i.e., 10 belongs to the class interval 10-20 (and not to 0-10).
↪ The number of observations in each class is called Class frequency.
↪ It is assumed that the frequency of each class interval is centered around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class.
Class-mark = ^{(Upper limit + Lower limit)}/₂

Direct Method

The class marks serve as  x_i’s in this method. For the ith class interval, the frequency  f_i  corresponds to the class mark x_i. Now, the mean can be computed in the same manner as in case of ungrouped data.

͞x	=	fi xi
		fi

This method of finding the mean is known as the Direct Method.
↪ This method gives an approximate mean because of the mid-point assumption.
↪ Remember, when this formula is used
(i) For Ungrouped frequency distribution,
x_i = ith observation
f_i = frequency of the ith observation.
(ii) For Grouped frequency distribution,
x_i = class mark of the ith class interval
f_i = frequency of the ith class interval. 

Ex - Find the mean for given data

Solution - We can write the given data in grouped frequency distribution table as following

So, the mean x̄ of the given data is given by

͞x	=	fi xi
		fi

= ¹⁸⁶⁰/₃₀ = 62

Assumed Mean Method

Sometimes when the numerical values of x_i (class mark) and f_i are large, finding the product of x_i and f_i becomes tedious and time consuming. We can't change the f_i’s, but we can change each x_i to a smaller number, so that our calculations become easy. We can achieve this by subtracting a fixed number from each of these x_i’s.
↪ The first step is to choose one among the x_i’s as the assumed mean, and denote it by ‘a’. We may take ‘a’ to be that x_i which lies in the center of x₁, x₂, . . ., x_n.
So, in previous example, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5.
↪ The next step is to find the difference between a and each of the x_i’s, that is, the deviation (d_i) of ‘a’ from each of the x_i’s i.e.,
d_i = x_i – a 
= x_i – 47.5
↪ The third step is to find the product of d_i with the corresponding f_i, and take the sum of all the f_id_i’s (Σf_id_i).
↪ Then the mean of the deviations,

͞d	=	fi di
		fi

↪ Since in obtaining d_i, we subtracted ‘a’ from each x_i, so, in order to get the mean ͞x , we need to add ‘a’ to d . This can be explained mathematically as:
Mean of deviations,

͞d	=	fi di
		fi

	=	fi (xi -a)
		fi

	=	fi xi		-		fi a
		fi				fi

͞d	=			͞x - a		fi
						fi

͞x = a + ͞d

͞x	=			a +		fidi
						fi

∴ Mean = Assumed Mean + Mean of deviations
For previous example, we can write mean deviation table as following (a = 47.5)

Substituting the values of a, Σf_id_i and Σf_i from Table, we get
͞x  = 47.5 + 435/30
47.5 + 14.5 = 62
Therefore, the mean of the marks obtained by the students is 62.

Step-deviation method

↪ In previous example, if we find the mean by taking each of x_i (i.e., 17.5, 32.5,and so on) as ‘a’, then the mean determined in each case is the same, i.e., 62.
So, we can say that the the value of the mean obtained does not depend on the choice of ‘a’.
↪ We can also observe that deviations are common multiples of the class size i.e., the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with f_i. (Here, 15 is the class size of each class interval.)
↪ Let 

ui	=	xi −a
		h

where a is the assumed mean and h is the class size.
↪ Then, Mean of reduced deviations,

͞u	=	fi ui
		fi

↪ Now, ͞x  can be find as following

͞u	=	fi ui
		fi

	=	fi (xi -a)/h
		fi

hū = x̄ - a
x̄ = a + hū

x̄ =	a + h		fi ui
			fi

↪ For previous example, we can write the step deviation table as follow (a = 47.5)

Now, substituting the values of a, h, Σf_iu_i and Σf_i from the Table, we get
x̄ = 47.5 + 15ײ⁹/₃₀
= 47.5 + 14.5 = 62
So, the mean marks obtained by a student is 62.
The method discussed above is called the Step-deviation method.

Note :
↪ the step-deviation method will be convenient to apply if all the d_i’s have a common factor (=h).
↪ The mean obtained by all the three methods is the same (an approximate mean).
↪ The assumed mean method and step-deviation method are just simplified forms of the direct method. Calculation is simplified by reducing x_i .
↪  The choice of method to be used depends on the numerical values of x_i and f_i. If x_i and f_i are sufficiently small, then the direct method is an appropriate choice. If x_i and f_i are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and x_i are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the d_i’s.
↪ The formula x̄ = a + hū still holds if a and h are not as given above (i.e., a = x_i & h = class size), but are any non-zero numbers such that u_i = ^{(x_i − a)}/_h .

Measure of Central Tendency for Ungrouped Data
	Mode & Median for Grouped Data ➤