2 CONTENTS OF THE CHAPTER 1. MeanBy direct methodBy assumed mean methodBy step-deviation method2. MedianOf grouped dataOf ungrouped data3. Mode
3 4. Graphical representation of cumulative frequency distribution Less than type cumulative frequency distributionMore than type frequency distributionAlso finding the missing frequencies when Mean is given, when Median is given etc.
4 Median: According to CANNOR, Mean: Mean is the number which is obtained by adding the values of all the items of a series and dividing the total by the number of items.Median: According to CANNOR,“The median is that value of the variable which divides the group into two equal parts,one part comprising all values greater than the median and the other part comprising all the values smaller than the median value.”3) Mode: According to KENNY,“The value of the variable which occurs most frequently in a distribution is called the mode.”
5 Mean of raw dataIf xı, x2, xз xn are n values of a variable X, then the arithmetic mean of these values is defined as:Mean= x1+x2+x xn / nIn other words, we can say that the Arithmetic mean of a set of observations is equal to their sum divided by total number of observations.
6 Example:The first 10 students of class 10TH A scored the following marks in 2nd unit test.35,30,32,40,35,40,25,20,25,30Find the mean.Here x1=35,x2=30, x3= x10=30Mean=x1+x2+x x10 / 10Mean= / 10=312 / 10 =31.2Average= mean marks= 31.2
7 In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise same method to find its mean.
8 Example:The marks obtained by 30 students of class 10TH of a certain school in mathematics paper consisting of 100 marks are presented below. Find out the mean of the marks obtained by the students.Marks obtained(x1) no. of students(f1)
9 Solution :To find mean marks, we require the product of each x1 with the corresponding frequency f1. So, let us put them in a column as shown:
11 NowMean= Σf1x1 = 1779 = 59.3Σftherefore, the mean marks obtained is 59.3.
12 The following algorithm may be used to compute Arithmetic mean by DIRECT METHOD: Prepare the frequency table.Multiply the frequency of each row with the corresponding values of variable to obtain third column containing f1x1.Find the sum of all entries in column 3 to obtain Σf1x1.Find the sum of all the frequencies in column 2 to obtain Σf1= N.Use the formula Mean= Σf1x1N
13 Now let us convert the data taken in the previous example by forming class-intervals of width say 15.Remember that, while calculating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class.Example: 4 students who have obtained 40 marks would be considered in the class-interval of and not in With this convention in our mind, let us form a grouped frequency distribution table as under
14 Class interval no. of students Now for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each class-interval is centered around its mid-points.
15 So,we find the mid-point of a class(or its class mark) by finding the average of its upper and lower limits.That is class mark= upper limit+lower limit2i.e. for the class 10-25, the class mark is =17.5Similarly we can find class marks of the remaining class-intervals. These class marks serve as our x1.Now we can proceed to compute the mean in the same manner.
16 class-interval no. of stds class mark f1x1 Σf1= Σf1x1=1860Mean =Σf1x1 = 1860 = 62.Σf
17 We observed that of the example 2 & 3 are using the same data and employing the same formula for the calculation of the mean but the result obtained are different.Can you think WHY THIS IS SO and WHICH ONE IS MORE ACCURATE?The difference in the two values is because of the mid-point assumption in the example 3.59.3 being the exact mean, while 62 an approximate mean.