
 Mean
 Mean Of Grouped Data
 Class Limits in Exclusive and Inclusive Form
 construct a pie chart
 Frequency Distribution of Ungrouped and Grouped Data
 Mode
 Real Life Statistics
 Mean of the Tabulated Data
 Median
 Construction of Bar Graphs
 Properties of Arithmetic Mean
 Average
 Word Problems on Average
Mean or average or arithmetic mean is one of the representative values of data. We can find the mean of observations by dividing the sum of all the observations by the total number of observations
Mean of raw data:
If x_{1}, x_{2}, x_{3}, ......x_{n} are n observations, then
Arithmetic Mean =` (x_1, x_2, x_3, ......x_n)/n`
= `[sum(x_i)]/n`
`sum` (Sigma) is a Greek letter showing summation
1. Weights of 6 boys in a group are 63, 57, 39, 41, 45, 45. Find the mean weight.
Solution:
Number of observations = 6
Sum of all the observations = 63 + 57 + 39 + 41 + 45 + 45
= 290
Therefore, arithmetic mean = `(290)/6` = 48.3
Mean of tabulated data:
If x_{1}, x_{2}, x_{3}, ....x_{n}are n observations, and f_{1}, f_{2}, f_{3}, .....f_{n} represent frequency of n observations.
Then mean of the tabulated data is given by
= `(f_1 x_1 + f_2 x_2 + f_3 x_3 + .... f_n x_n)/(f_1 + f_2 + f_3 + ....f_n)`
= `[sum(f_ix_i)]/[sum f_i] `
2. A die is thrown 20 times and the following scores were recorded 6, 3, 2, 4, 5, 5, 6, 1, 3, 3, 5, 6, 6, 1, 3, 3, 5, 6, 6, 2.
Prepare the frequency table of scores on the upper face of the die and find the mean score.
Number on the upper face of die  Number of times it occurs (frequency)  f_{i}x_{i} 

1  2  1 `xx`2 = 2 
2  2  2 `xx` 2 = 4 
3  5  3`xx` 5 = 15 
4  1  4 `xx` 1 = 4 
5  4  5 `xx` 4 = 20 
6  6  6` xx` 6 = 36 
Therefore, mean of the data = `[sum(f_ix_i)]/[sum f_i] `
= `(2 + 4 + 15 + 4 + 20 + 36)/(20)`
= `(81)/(20)`
= 4.05
Mean of the data=`[sum(f_ix_i)]/[sum f_i] `
3. If the mean of the following distribution is 9, find the value of p.
X  4  6  p + 7  10  15 
f  5  10  10  7  8 
Solution:
Calculation of mean
x_{i}  f_{i}  x_{i}f_{i} 

4  5  20 
6  10  60 
p + 7  10  10(p + 7) 
10  7  70 
15  8  120 
∑f_{i} = 5 + 10 + 10 + 7 + 8 = 40
∑ f_{i}x_{i} = 270 + 10(p + 7)
Mean =`[sum(f_ix_i)]/[sum f_i] `
9 = `(270 + 10(p + 7))/(40)`
270 + 10p + 70 = 9 `xx` 40
340 +10p = 360
10p = 360  340
10p = 20
p = `(20)/(10)`
p = 2
While calculating the mean of the grouped data, the values x_{1}, x_{2}, x_{3}, ....x_{n} are taken as the midvalues or the class marks of various class intervals. If the frequency distribution is inclusive, then it should be first converted to exclusive distribution.
4. The following table shows the number of plants in 20 houses in a group.Find the mean number of plans per house
Number of Plants  (0  2)  (2  4)  (4  6)  (6  8)  (8  10)  (10  12)  (12  14) 
Number of Houses  1  2  2  4  6  2  3 
Solution:
We have
Number of Plant  Number of Houses (f_{i})  Class Mark (x_{i})  f_{i}x_{i} 
(0  2)  1  1  `(1 xx 1)` = 1 
(2  4)  2  3  `(2 xx 3)`= 6 
(4  6)  2  5  `(2 xx 5)`= 10 
(6  8)  4  7  `(4 xx 7)`= 28 
(8  10)  6  9  `(6 xx 9)` = 54 
(10  12)  2  11  `(2 xx 11)` = 22 
(12 14)  3  13  `(3 xx 13)` = 39 
∑f_{i} = 1 + 2 + 2 + 4 + 6 + 2 + 3 = 20
∑f_{i}x_{i} =1 + 6 + 10 + 28 + 54 + 22 + 39 = 160
Therefore, mean =`[sum(f_ix_i)]/[sum f_i] `
= `(160)/(20) = 8` plants
In class boundaries and class limits in exclusive and inclusive form we will mainly discuss about;
 Class limits and class boundaries in exclusive and inclusive forms
 Class interval, class mark, range of the data
Class limits in exclusive and inclusive form:

In exclusive form, the lower and upper limits are known as true lower limit
and true upper limit of the class interval.
Thus, class limits of 10  20 class intervals in the exclusive form are 10 and 20.  In inclusive form, class limits are obtained by subtracting 0.5 from lower limit
and adding 0.5 to the upper limit.
Thus, class limits of 10 20 class interval in the inclusive form are 9.5  20.5.  Class size: Difference between the true upper limit and true lower limit of a class interval is called the class size. Class size remains the same for all class intervals. For the class interval 10  20 Class size is 10, i.e., (20 10 = 10)
 Class mark: Midvalue of each class interval is called its class mark. Class mark = `1/2` (Upper limit + Lower limit) For the class interval 10  20 Class mark = `(10 + 20)/2` =`(30)/2` = 15

Range: The difference between the maximum value and the minimum value of the observation is called the range.
In the above data, the maximum value is 24 and the minimum value is 0.
Therefore, range = 24  0 = 24
Frequency distribution of ungrouped data:
Given below are marks obtainedby 20 students in Maths out of 25. 21, 23, 19, 17, 12, 15, 15, 17, 17, 19, 23, 23, 21, 23, 25, 25, 21, 19, 19, 19
Frequency distribution of grouped data:
 The presentation of the above data can be expressed into groups. These groups are called classes or the class interval.
 Each class interval is bounded by two figures called the class limits.
Marks  Number of Students (Frequency) 

(0  10)  0 
(10  20)  11 
(20  30)  9 
Exclusive form of data
 This above table is expressed in the exclusive form.
 In this, the class intervals are 0  10, 10  20, 20  30. In this, we include lower limit but exclude upper limit.
 So, 10  20 means values from 10 and more but less than 20.
 20 30 would mean values from 20 and more but less than 30.

Data in the inclusive form:
Marks obtained by 20 students of class VIII in Maths text are given below.  23, 0, 14, 10, 15, 3, 8, 16, 18, 20, 1, 3, 20, 23, 24, 15, 24, 22, 14, 13
Let us represent this data in the inclusive form.
Marks  Number of Students (Frequency) 

(0  10)  6 
(11  20)  9 
(21  30)  5 
 Here, also we arrange the data into different groups called class intervals, i.e., 0  10, 11  20, 21 30.
 0 to 10 means between 0 and 10 including 0 and 10.
 Here, 0 is the lower limit and 10 is the upper limit. 11 to 20 means between 11 and 20 including 11 and 20.
 Here, 11 is the lower limit and 20 is the upper limit.
 When the data is expressed in the inclusive form, it is converted to exclusive form by subtracting 0.5 from lower limit and adding it to upper limit of each class interval.
 11  20 is expressed in the inclusive form which can be changed and taken as 10.5  20.5 which is the exclusive form of the data.
 Similarly, 21 30 can be taken as 20.5  30.5.
For a set of given observation, mode is that observation which occurs maximum number of times.
1. Find the mode of the given set of number
2, 2, 3, 5, 4, 3, 2, 3, 3, 5
Solution:
Arranging the number with same values together, we get
2, 2, 2, 3, 3, 3, 3, 4, 5, 5
We observe that 3 occurs maximum number of times, i.e., four.
Therefore, mode of this data is 3.
Note: A data may not have a mode.
2. The data 3, 4, 1, 5, 4, 2 has no mode because no number occurs more number of times then any other number.
A data may have more than one mode.
3. The data 2, 5, 1, 3, 5, 7, 6, 3, 8 have two modes 3 and 5.
Therefore, each is repeated two times which is maximum
4. The height of 50 plants in a garden are given below. Find the mode of the data.
Height (cm)  10  25  30  40  45 
Number Plants  13  15  12  8  2 
Solution:
The frequency of 25 is maximum.
So, the mode of this data is 15.
 In real life statistics, we come across numerical data in the newspapers, magazines and television regarding different aspects like increase or decrease in population, profit made by a company in different years, weather report, etc.
 These numerical facts are also represented by graphs which are easy to understand.
Definition of statistics:
 The term statistics is derived from a Latin word status meaning condition. The branch of mathematics which deals with the collection, presentation, analysis and interpretation of the numerical data is called statistics.
 We have already learned how to draw and read the pie graph, column graph and line graph in the statistics sixth grade. Now, we will learn the terms related to statistics to handle the raw data, make frequency distribution table and calculate mean.
In line graph:
 Points are plotted on the graph related to two variables
 Points are joined by the line segments.
How to Construct a Line Graph?
Steps of construction of line graph:
 On a graph, draw two lines perpendicular to each other intersecting at O. The horizontal line is xaxis and vertical line is yaxis.
 Mark points at equal intervals along xaxis and write the names of the data items whose values are to be marked.
 Along the yaxis, choose an appropriate scale considering the given values.
 Now, make the points Join each point with the successive point using a ruler. Thus, a line graph is obtained
1. The following table gives the information of the sum scored by McKay in 10 matches. Represent this information using line graph.
Match  1  2  3  4  5  6  7  8  9  10 
Runs Scored  80  30  50  90  40  60  75  20  15  90 
2. Mobile phones sold by a shop in a certain week are as follows:
Days  Mon  Tues  Wed  Thurs  Fri  Sat 
No. of Mobile Phones Sold  20  16  12  24  4  10 
Represent the above data by a line graph.
In the mean of the tabulated data, if the frequencies of n observations x_{1}, x_{2}, x_{3}, .... x_{n} are f_{1}, f_{2}, f_{3} ...... f_{n}, then Mean of the tabulated data = `((f_1x_1 + f_2x_2 + f_3x_3 ..... f_nx_n))/(( f_1 + f_2 + f_3 .... f_n)) = (sum(f_ix_i))/(sum f_i) `
1. Find the mean weight of 50 girls from the following table
Weight in kg  40  42  34  36  46 
No.of girls  6  6  15  14  7 
Solution:
Mean = `(f_1x_1 + f_2x_2 + f_3x_3+ f_4x_4 + f_5x_5)/(f_1 + f_2 + f_3 + f_4 + f_5)`
= `((40 xx 6) + (42 xx 6 )+ (34xx 15 )+ (36 xx 14) + (46 xx 7))/(8 + 6 + 15 + 14 + 7)`
= `(240 + 252 + 510 + 504 + 322)/(50)`
= `(1828)/(50)`
= 36.56
2. If the mean of the following frequency distributions is 9, find the value of 'a'. Write the tally marks also.
Variable (x_{i})  4  6  8  10  12  15 
Frequency (f_{i})  8  9  17  a  8  4 
Solution:
Frequency distribution table
But given mean = 9
So, we have `(378 + 10a)/(46 + a)` = 9
378 + 10a = 9(46 + a)
378 + 10a = 414 + 9a
10a  9a = 414  378
a = 36
 Median of a group of observation is the value which lies in the middle of the data (when arranged in an ascending or descending order) with half of the observations above it and the other half below it.
 When the number of observations (n) is odd. Then, median is `[(n + 1)/2]^ (th)` observation.
 When the number of observations (n) is even. Then median is the mean of `(n/2)^(th)` and `(n/2+ 1)^(th) `observation.
i.e., Median =`{(n/2)^(th) observation +(n/2 + 1)^(th) observation}/2`
Let us observe the following solved problems using stepbystep explanation.
1. Find the median of the data 25, 37, 47, 18, 19, 26, 36.
Solution:
Arranging the data in ascending order, we get 18, 19, 25, 26, 36, 37, 47
Here, the number of observations is odd, i.e., 7.
Therefore, median = `((n + 1)/2)^(th)` observation.
= `((7 + 1)/2)^(th)`observation.
= `(8/2)^(th)` observation
= 4^{th} observation.
4^{th} observation is 26.
Therefore, median of the data is 26.
2. Find the median of the data 24, 33, 30, 22, 21, 25, 34, 27.
Solution:
Here, the number of observations is even, i.e., 8.
Arranging the data in ascending order, we get 21, 22, 24, 25, 27, 30, 33, 34
Therefore, median = `{(n/2)^(th) observation + (n/2 + 1)^(th) observation}/2`
= `(8/2)^(th)` observation + `(8/2 + 1)^(th)` observation
= 4^{th} observation + (4 + 1)^{th} observation
= `(25 + 27)/2`
= `(52)/2`
= 26
Therefore, the median of the given data is 26.
Bar graph is the simplest way to represent a data.
 In consists of rectangular bars of equal width.
 The space between the two consecutive bars must be the same.
 Bars can be marked both vertically and horizontally but normally we use vertical bars.
 The height of bar represents the frequency of the corresponding observation.
For example, let us observe the following data of the bar graph.
The following data gives the information of the number of children involved in different activities.
Activities  Dance  Music  Art  Cricket  Football 
No. of Children  30  40  25  20  53 
How to Construct a Bar Graph?
Steps in construction of bar graphs/column graph:
 On a graph, draw two lines perpendicular to each other, intersecting at 0.
 The horizontal line is xaxis and vertical line is yaxis.
 Along the horizontal axis, choose the uniform width of bars and uniform gap between the bars and write the names of the data items whose values are to be marked.
 Along the vertical axis, choose a suitable scale in order to determine the heights of the bars for the given values. (Frequency is taken along yaxis).
 Calculate the heights of the bars according to the scale chosen and draw the bars.
 Bar graph gives the information of the number of children involved in different activities.
1. The percentage of total income spent under various heads by a family is given below.
Different needs  Food  Clothing  Health  Education  House Rent  Miscellaneous 
% Age of Total Number  40%  10%  10%  15%  20%  5% 
2. 150 students of class VI have popular school subjects as given below:
Subject  French  English  Maths  Geography  Science 
Number of Students  30  20  26  38  34 
Draw the column graph/bar graph representing the above data.
Solution:
Take the subjects along xaxis, and the number of students along yaxis
Bar graph gives the information of favourite subjects of 150 students.
3. The vehicular traffic at a busy road crossing in a particular place was recorded on a particular day from 6am to 2 pm and the data was rounded off to the nearest tens.
Time in Hours  (6  7)  (7  8)  (8  9)  (9  10)  (10  11)  (11  12)  (12  1)  (1  2) 
Number of Vehicles  100  450  1250  1050  750  600  550  200 
Bar graph gives the information of number of vehicles passing through the crossing during different intervals of time.