Data Handling

Data Handling -How to read Pie chart, Bar Graph, Line Graph

Average (Mean): 

Concept of average: Let's say we have to compare two sections of a class on the basis of their numbers in mathematics. There are many students in each class and it is not feasible to compare each student in one class with another. The easy way is to calculate the average marks scored by both the section and then compare the average data.

Let there are total n entities and their values be x_1, x_2, x_3,... x_n 

Then , their average is defined as 

Average =     ( x₁   +   x₂  + x₃ + ...+ x_n) / n

A student has scored 52, 85, 79, 65, 64 marks in 5 subjects, then 

their average will be defined as 

Average or Mean = ( 52 + 85 + 79 + 65 + 64)/5

                          = 69   

Median:   A mean is defined for a group of data. It is a value that lies in mid of data values arranged in either ascending or descending order.

The rule of calculating the median is :

1. First arrange the numbers in increasing or decreasing order.

2. 
    (a)If the number of items is odd, then just divide the number of items by 2 and keep the integral part, say "m". The value just after the mth item is the median of the series

    (b) If the number of items in the series is even then just divide the number of items by 2. The median will be the average of the value at mth place and (m+1)th place. 

Examples: 

1. Number of items in the series is odd:

   Taking the same example that was used for calculating average. A student's    

   scored marks in 5 subjects are 52, 85, 79, 65, 64. 

            Arranging this series in increasing order, we have 52, 64, 65, 79, 85. 

   Now, there are five values and the mid value is 65. It has 2 values on the 

   left and 2 values on the right. Hence it lies in the mid.

          Therefore median would come out to be 65.

2. Number of items in the series is even

    Lets say we add one more item in the above series, 95. Then the series would  be 52, 64, 65, 79, 85, 95. This series has 6 items. An even number, therefore the median will be the average of 6/2 = 3rd value and (3 +1)th value i.e. (65 + 79) /2 = 72

Mode: Mode is a number that repeats maximum times in the series.

Example: Lets say the series is 4, 5, 9, 1, 5, 6, 7, 8,5

Arrange the series in increasing order.We have,

                

                  1,4,5,5,5,6,7,8,9 

We can see that the number 5 repeats maximum times. Hence, 5 is the mode of the series.

Range : Range is the difference between the highest and the lowest values.

In our example above for mode, the highest value is 9 and the lowest value is 1 and therefore the range is (9 - 1) = 8.

Pie -Chart :

Pie-chart is a representation of data using a circle and its sectors to represent the breakup of the data. Magnitude of the data is translated to the degrees in the circle.

  A Sample pie chart is given below:

Bar Graph:  A bar graph represents the data using rectangle ( known as bars) and the length of the bar represents the magnitude.

Solved Questions:

Q1. The mean of the median, mode and range of the observations
               6, 6, 9,14, 8, 9,9,8
is 
(1) 8.5
(2) 8.8
(3) 10.3
(4) 10.5

Answer: (1) 8.5
Solution:
Mode = 9
Median = 8.5
Range = Highest value - lowest value = 14 - 6 = 8
Taking total of all the above values = 8.5 + 9 + 8 = 25.5
Therefore, mean = 25.5/3 = 8.5

Q2:

Answer: (2) April - May

3.  The bar graph provided below gives the data of the production of paper ( in lakh tonnes) by three different companies X, Y and Z  over the years. Study the graph and the answer the questions that follow:

Production of Paper ( in lakh tonnes) by three companies x, Y, Z over the years

.

Bar graph

(i).  What is the difference between the production of company Z in 1998 and company Y in 1996?

   (a) 2,00,000 tons (b) 20,00,000 tons 

   (c) 20,000           (d) 2,00,00,000 tons

   (e) None of these

(ii). What is the ratio of the average production of company X in the period 1998-2000 to the average production of Company Y in the same period?

    (a) 1:1     (b) 15:17 

    (c) 23:25  (d) 27:29

    (e) None of these

(iii).  What is the percentage increase in the production of Company Y from 1996 to 1999?

    (a) 30%   (b) 45% 

    (c) 50%   (d) 60% 

    (e) 75%   

(iv). The average production for five years was maximum for which company ?

    (a) X    (b) Y

    (c)  Z   (d) X and Y

    (e) X and Z both

(v). For which of the following years, the percentage rise/fall in the production from the previous year is the maximum for company Y?

    (a) 1997   (b) 1998 

    (c) 1999   (d) 2000 

    (e) 1997 and 2000

(vi). In which year was the percentage of production of company Z to the production of company Y the maximum?

    (a) 1996  (b) 1997

    (c) 1998   (d) 1999 

    (e) 2000

Solution : 

(i). (b)

     Production of company Z in 1998  =  45 

     Production of company Y in 1998  =  25 

     Hence, difference =            45 -25 =  20      

(ii) (c)

     Average production of company X in the period 1998-2000 =  115/3

     Average production of company Y in the period 1998-2000 =  125/3

     Hence, there ratio is                                                              =  23:25

    

(iii)  (d)

    Production of company Y in 1996  = 25

    Production of company Y in 1999  = 40

    Therefore , production increase = (15/ 25)x100 = 60%

(iv) (a)

     Rise in production for company y in year 1997  =  (10/25)X100 = 40%

     Rise in production for company y in year 1998  =  (0/35)X100   =   0%

     Rise in production for company y in year 1999  =  (5/35)X100   =  14.28 %

     Rise in production for company y in year 1998  =  (10/40)X100   = 25 %

     Hence, rise was maximum in the year 1997

(v) (a)

     In the year 1996 production of company Z to company Y

                               = (35/25)x100 = 140%

     In the year 1997 production of company Z to company Y

                               = (40/35)x100 = 114.28%

     In the year 1998 production of company Z to company Y 

                               = (45/35)x100 = 128.57%

     In the year 1999 production of company Z to company Y

                               = (35/40)x100 = 87.5%

     In the year 200 production of company Z to company Y 

                               =   (35/50)x100 = 70%

4. The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India ( NHAI) for its phase II projects. Study the Pie-Chart and Answer the questions that follow.

Pie Chart

Total funds to be arranged for projects ( Phase II) = Rs. 57,600 crores

Q1. Near about 20 % of the funds are to be arranged through :

   (a) SPVS (b) External Assistance (c) Annuity (d) Market Borrowing

Q2. The central angle corresponding to Market Borrowing is :

(a) 52           (b)   137.8         (c) 187.2          (d) 192.4

Q3. The approximate ratio of the funds to be arranged through Toll and that through Market Borrowing is :

(a) 2 :9        (b)  1:6        (c) 3:11             (d) 2:5

Q4. If  NHAI could receive a total of Rs. 9695 crores as External Assistance, by what percent ( approximately) should it increase the Market Borrowing to arrange for the shortage of funds:

(a) 4.5%        (b) 7.5 %        (c)  6%            (d) 8%

Q5. If the Toll to be collected through an outsourced agency by allowing a maximum 10 % commission, how much amount should be permitted to be collected by the outsourced agency, so that the project is supported with Rs. 4910 crores.?

(a) Rs. 6213 crores  (b) Rs. 5827 Crores         (c)Rs. 5401 Crores             (d) Rs.  5216 Crores

Solution:

(i). (b)External Assistance

         20 % of 57600 crore would be 11520 crores that is near to (b) External Assistance

(ii). (c)

      Use the formula = (Value of one / Value of total) x 100

      we wil get (29952/57600) x 100 = 187.2

(iii). (b)

         Toll share nearly equal 5000 and Market borrowing nearly equal 30,000. Dividing them to get 1/6.

(iv). (c) 

       If they can get 9695 as external assistance then they will be short by  ( 11486 - 9695 = 1791 )  and the 

      percentage market borrowing should be raised to get this amount is (1791/29952)x100 = 6%

(v).  (c)  

        Amount required to support the project with 4910 allowing the outsourcing agency to collect 10 % is = 

       4910 + (.1) x 4910 = 4910 + 491  = 5401