15 Representation and Interpretation of Data

Transcription

1 15 Representation and Interpretation of Data After studying this lesson you will acquire knowledge about the following : Drawing a histogram for data with unequal class intervals Drawing a frequency Polygon using the histogram Drawing a Cumulative frequency curve Finding quartiles and the inter quartile range Representing data by graphs 15.1 Histogram of a frequency distribution with equal class intervals You have learnt before how to draw the histogram for a frequence distribution with equal class intervals. To revise it, consider the following example first. Example 1 The marks obtained bythe students in aclass fortest, out of50 isshown by the frequency distribution below. Drawa histogram forit:here0 < xx < 10 Marks No:of students (frequency) The histogram for this frequency distribution is as follows. 12 > Number of Students > Marks

2 Exercise 15.1 The width of thecolumns areequalin this histogram. The frequency of the class interval is represented by the height of a column. The area of the rectangle relevant to each class interval is proportional to the frequency. (1) The time taken by each athlete to finish a race is given below. Draw a histogram to represent this information:here35 < x < 45 Time (to nearest minute) Frequency No. of athletes (2) The data obtained by measuring weights of students in a class on a certain day are denoted in the following distribution. Draw a histogram for this data. Weight (W) Kg Frequency (No. of Students) (3) The informations about the weight of several players is given in the table below. Represent these data in a histogram. Wt (Kg) No.of Players (Frequency) 173

3 (4) The number ofpackets ofa new brand ofice cream,sold duringthe first 200 days, afterintroducing itto themarket, is given below in the frequency table. Representthis in a histogram. No.of packets sold Frequency (No. of dates) (5) The data obtained by measuring circumference of trees in a garden are given in the table below. Circumference (cm) Frequency (No of dates) (i) Find the modal class. (ii) Represent this data in a histogram. (iii) According to the histogram what is the modal class? 15.2 The histogram for data with unequal class intervals Example 2 A Company manufactures metal rods in various lengths. The table below shows information of a day s production. length (cm) No.of bars (Frequency) Here the size of the first four intervals is equal but the 5 th is 50-70, 6 th is and the 7 th is They are not of the same size. 174

4 As the size of the first fourintervals is 10, thearea ofthe firstfour rectangles are 40,60,80,100 respectively. But the size of the interval is 20. We have to find the height ofthis rectangleso as to havean areaof 100square units. 100 ie. the height of this rectangle 5 20 If you consider the size 10 of the class interval as one unit, the height of the rectangle corresponding to the class can be obtained by dividing the frequency by 2 ie Therefore for the given distribution the heights of the rectangles should be adjusted as shown below. Length No. of rods Height of the cm (f) rectangles According to the above table, the histogram is as follows. Number of rods Length (cm) 175

5 Example 3 Draw a histogram to represent the data given below about the heights of a group of students. Height (cm) Frequency Height of (No.of students) Rectangles See how the heights of the rectangles are calculated taking the width of a class 5 cm as one unit. Accordingly the width of the class is 2 units. For the area to be proportional to the frequency, the frequency 18, when divided by 2, gives the height of the rectangle, 9. Similarly for the remaining classes where the class intervals are not the same, observe how the heights of the rectangles are calculated. The corresponding histogram is shown below. > 12 Number of Students height (cm) > 176

6 So far we here considered drawinghistograms for continuous data. Now Consider the frequency distribution given below. Example 4 The table below shows the amount of telephone calls taken by 100 institutes per day. Number of Telephone calls Number of Institutes These are called descrete data. Therefore we cannot draw a histogram for this data as discussed earlier. So the histogram should be drawn by organizing this data as continuous data using class boundaries and calculating the hights of rectangles. No. of Telephone Class No. of height of the Calls boundaries Institutes rectangles

7 Number of Institutes Histogram Number of Telephone calls Exercise 15.2 (1) The table shows the results obtained from an investigation done regarding the life span of a certain brand of electric bulbs. Duration Period (Hrs) Class intervals (Number of Light Bulbs) (2) The table shows the marks obtained by a group of students in an examination. By taking the class boundaries draw a histogram for this data. Class Intervals (Marks) Frequency (Number of students) 178

8 (3)A histogram representing information about the weight of a fish is shown below. No. of fish weight of one fish in kg (i) How many fish are there weighing from 6 to 8 kg? (ii) How many fish are there weighing from 6 to 10 kg? (iii) What is the total number of fish weighed to obtain this data? 15.3 The Frequency Polygon The Frequency Polygon can be used to represent data as well as the histogram. There are two ways of constructing a frequency Polygon. 1. Using histogram 2. Using the mid value of each class interval and the relevant frequency When the class intervals are equal, the frequency polygon can be obtained by drawing straight line segments, joining the mid point of the class interval prior to the 1 st class interval of the histogram and the mid point of the top of each rectangle then the mid point of the class interval next to the last class interval Let us consider the infomation about the marks of an examination in example 1. Marks No. of Students (Frequency)

9 Let us draw a frequency polygon to represent this data. For this we have to draw the histogram first. Number of Students Marks Now the mid point A, of the class interval prior to the first class interval ie. 0-10, is joined to the mid point B of the 1 st column. In the same manner join B,C,D,E,F and finally join the mid point G of the class interval after the last class interval, and complete the polygon Now what is obtained is the frequency polygon. A frequency Polygon has follwing properties The area of the frequency Polygon is equal to the area of the histogram. (Considering the triangles CYX and PDY it is realised that the area removed from the histogram is equal to the area added to the frequency Polygon.) One side of the polygon is the x - axis Without drawing a histogram, the frequency polygon can be drawn by taking the mid value of each class interval and the corresponding frequency as an ordered pair and by plotting and joining the points in the correct order. 180

10 Number of Students The frequency polygon can be obtained by joining the points A, (5,3), (15,8), (25,12), (35,6), (45,5) and G respectively The frequency Polygon for a frequency distribution with unequal class intervals Example 4 Let us draw a frequency polygon using the histogram for the frequency distribution in example 2 Method I Marks Number of rods Length (cm) 181

11 The frequencypolygon canbe drawn in the same wayby separatingthe classes ofunequal sizes to equalsizes. It is shown in thediagram above. Method II Number of rods Length (cm) It can also be drawn by joining the mid point of the difference between the heights of the columns. As the first four classes have the same size from S to A the line can be drawn as before.then join A to the mid point X of BY and draw the line AXC. Now join C to the mid point D of PQ and draw the line CDE. Similarly complete the rest of the polygon. Veryfy whether the parts added to the polygon and the parts removed have the same area. Exercise 15.3 (1) The table below gives the information about the ages of a group of children in a play ground. AGE (Years) Frequency (No:of Children) (i) Draw a histogram representing these data. (ii) Draw the frequency Polygon with the use of the histogram. (iii) What can you say about the areas of the histogram and the frequency Polygon. 182

12 (2) The information about the heights of 80 students in aschool is denoted below. Class intervals (height cm) Frequency (No of students) (i) Draw a histogram to represent the above data and usingit construct the frequency polygon. (3) A histogram drawn to represent the infomation about ages of Patients who had come to a clinic to get treatment is shown below. Number of patients age (i) If there were 5 patients between find the number of Patients relevent to each age according to the histogram. (ii) Draw a frequency polygen using the histogram. (4) The information obtained from a survey about ages of farmers in a village are given below. Age in years Frequency (No of students)

13 (i) Considering the class boundaries, draw a histogram to represent this data. (ii) Draw a frequency Polygon usingthe histogram Cumulative frequency Curve and Quartiles Look at the table below. It gives infomation about the time taken for a telephone call by each caller at a post office during a day. Class Interval Frequency Cumulative time (minutes) (number of Frequency customers) Each value in the cumulative frequencey column is obtained by adding the frequency in that row and all the frequencies in the rows above it. The frequency thus obtained is called the cumluative frequency. The table with cumulative fequencies is called a cumulative fequency table. Considering the first and the third columns of the table, the number of customers who had taken less than 2 minutes is 4 and those who have taken less than 4 minutes. is 10. When they are written as orderd pairs (2,4), (4,10), (6,20), (8,28), (10,31), (12,32) are obtained. By plotting these points on a co-ordinate plane and joining them, a smooth curve as shown below is obtained. 184

14 Cumulative frequency This is known as the Cumulative frequency curve Time (minutes) In the graph, look at the horizontal line denoted by A. This is drawn through 16, which is exactly half the total number of callers. The x co-ordinate of the point where this line meets the curve is What is this value? You already know that the value correspoing to the middle score is the median. Hence according to the given information, the median is 5.25 mnts. Now look at the horizontal lines B and C. B is drawn through the mid value of the first half. ie through 8 which is the 1 of 32. The x co-ordinate of the 4 point where this line meets the curve is This is called the first quartile and is denoted by Q 1 Q The horizontal line C is drawn through the mid value of the second half. ie. Throught 24 which is 3 4 of 32 The x co-ordinate of the point where this line meets the curve is 7. This is called the third quartile and is denoted by Q 3 Q 3 = 7 mnts The value Q 3 - Q 1 is known as the interquartile range Here Q 3 - Q 1 = mnts. = 3.75 mnts Interquartile range= 3.75mnts. 185

15 Example 5 A Person maintaining a shop recorded his daily income for 30days. The result is shown in the table. Income (Rupees) No. of days (i) Draw a Cumulative frequency curve for the above data. (ii) Find the median income per day. (iii) Find the inter quartile range. Class Interval Frequency Cumulative (Income - Rupees) (days) Frequency According to the table above the cumulative frequency curve can be drawn as shown below (i) Cumulative Frequency (Income- Rupees)

16 (ii) According to the graph the median income - Rs 360/= (iii) 1 st quartile (Q 1 ) = Rs 240/- 3 rd quartile (Q 3 ) = Rs 460/- Exercise 15.4 The inter quartile range = Q 3 - Q 1 = Rs ( ) = Rs 220/= (1) The number of shirts completed within a week by a group of girls in a garment factory is given below. No of shirts Frequency (No of girls) (i) Find the first, second and the third quartiles of the number of shirt completed within a week. (ii) Find the inter quartile range. (2) Given below are the number of days absent from school during one year by a group of students 12,9,10,11,12,0,11,2,9,8,9,7,0,13,11 Find (i) First quartile (ii) Second quartile (iii) Third quartile (iv) Inter quartile range for this distribiution. (3) In a school, where there are parallel classes, the marks obtained by 240 students in grade 11 classes in a test are shown below. Marks Frquency

17 (i) If 25% of the students who obtained lowest marks are to be chosen what is the markthat should be taken as thelimit? (ii)if 25% of the students who got heighest marks are to be chosen what is the markthat should be taken as thelimit? (4) Information collected from a research done regarding a certain disease is given in the table. Age of patients (years) No: of Patients Find (i) (ii) Quartiles Inter quartile range of the distribution. (5) The table given below shows information about the temperature in a certain area Temperature C 0 No : of days (- 30) -(- 20) (- 20 ) -(- 10) (-10 ) (i) Calculate the annual median temperature of the area using this data. (ii) Find the inter quartile range. (6) Given below is the information about the lengths of pieces of ribbon Class intervals length of a piece in cm Frequency No. of pieces

18 (i) Prepare a cumulative frequency table. (ii) Draw a cumulative frequency curve and hence find the median and the interquartile range. (7) The time duration of calls taken from a post office by autside callers is given in the table below. Class interval (Time in seconds) frequency (No. of calls) (i) Prepare a cumulative frequency table. (ii) Draw a frequency curve and hence find the (a) mean (b) 1 st and the 3 rd quartiles (c) Interquartile range (8) In a certain school, the grade 6 admission test was done by 150 students. Information about the marks they obtained is given below. Class Interval (Marks) Frequency (No. of students) (i) Prepare a cumulative frequency table for this information. (ii) Draw a cumulative frequency curve and hence find the median of the distribution, first and the third quertiles and the interquartile range. 189

19 (9) Give below is the information about the salaries of employees in a certain institution.by drawing a cumulative frequency curve for this information, findthe medianand the interquartile range. Class Interval (salary-rupees) Frequency (No of employees) (10) In the year 1998, the information gathered from a survey done within 93 days about the price of a kilogram of beans is given in the table below. Class interval (Price in Rupees) Frequency (No. of days) By drawing a cumulative frequency curve to illustrate the above data find the median price of a kilogram of beans and the interquartile range. 190