ation of tabulated frequencies. This representation is done with the help of adjacent rectangles erected over discrete intervals.
Different Components of Histogram
Height of the Rectangular Bars It shows frequency density of the interval, i.e., the frequency divided by the
width of the interval.
Area Covered by the Bars This area shows the frequency of the observations in the interval.
Total Area of Histogram This area is equal to the number of data.
Directions (Examples 6-8) In the given histogram, there is survey for the different; family of different income group of a village. Study the graph carefully and answer the questions given below. ssc (1o+2) 2010
6. How many families were surveyed?
7. What is the total number of family whose daily income is Rs 800 or more?
8. What is the percentage of families whose daily income is less than Rs 500?
* If any bar has frequency zero, then no bar will be formed for that class interval.
* In histogram, the minimum and maximum value of interval of bar is called measurement of bars. Frequency density = Frequency of each class (bar)/Number of units of class interval
(b) Frequency Polygon
A frequency polygon shows a frequency distribution emphasizing the overall pattern in the data.
In this polygon fragmented categories are denoted an -Y-axis and frequencies are denoted on y-axis.
A frequency polygon can be constructed by two methods i.e., with the help of a histogram and without help of histogram.
I. Constructions of a Frequency Polygon with the help of Histogram
The frequency polygon is that type of polygon in which a closed two-dimensional figure of straight line segments joining the mid-points of the top of the bars of a histogram. The first point is located on the x-axis and is placed in the middle of the interval which precedes the first bar of the histogram.
The last point is located on the x-axis in the middle of the interval immediately following the last bar of the histogram.
Direction (Example 9) Draw a frequency polygon with the given data.
II. Construction of a Frequency Polygon without help of Histogram
To draw a frequency polygon with the help of given class intervals and frequency, first of all find the centre-point of each class interval, then indicate these point on x-axis with corresponding frequencies on y-axis. Now join the points according to parameters.
Direction (Example 10) Draw a frequency polygon with the given data.
Directions (Examples 11 -14) Given graph shows the yearly premium of an insurance company which is recovered for the insurance of Rs 1000 for different ages. Study the graph and answer the questions given below, ssc (10+2) 2012
11, Premium on Rs 10000 for the people of age 27 is
12. Two members of a family are of age 25 yr and 30 yr. What will be the premium for both of them for the insurance of Rs 10000 each?
(a) Rs 925
(b) Rs 910.60
(c) Rs 876.25
(d) Rs 885
13.If the person of age 28 yr is replaced by an another person of age 35 yr with an insurance of Rs 1000, then what will be per cent increase in the premium?
14. What will be the premium for a person of age 31 yr for the insurance of Rs 1000?
* The surface under the frequency polygon is exactly the same as the surface of the histogram i.e., it represents the same amount of data.
* Frequency polygon can be drawn with freehand. In frequency polygon, for any class, frequency of the unit class-interval is called frequency density.
Directions (Q.Nos. 1 -4) Study the table and answer the questions. ssc (10+2)2015
1.The number of years during which the company has its production less than the average production during 1920-1927 is approximately?
2. The average production of steel is
3. The difference of the production of steel in the year 1923 and 1924 is x% of 1927. Then, the value of xis approximately
4. The ratio of production of steel in the year 1924 and 1925 to that of 1923 and 1927 is
(a) 2005 : 2077
(b) 2205 : 2007
(c) 2205 : 2077
(d) 2077 : 2205
Directions (Q. NOS. 5-8) Study the graph and answer the questions. ssc (10+2) 2014
5. What was the pass percentage in year 2008?
6. What was the