1. SUBTOPIC 8.2 : Frequency Distributions

2. (a) Clarify the important terms in the construction of frequency table; class interval, class limit, class boundary, class width and class mark (b) Constructs frequency distribution table (c) Construct the Histogram and Frequency polygon OBJECTIVES (d) Construct a ‘less than’ cumulative frequency distribution table (e) Draw an ogive from the ‘less than’ cumulative frequency distribution table

3. FREQUENCY DISTRIBUTION The following are the number of leave of 30 lecturers of KMPP in the year 2007. 2, 3, 0, 0, 5, 1, 0, 0, 2, 0, 1, 4, 0, 1,1 5, 0, 0, 0,2, 1, 1, 3, 1, 0, 1, 1, 0, 2,0 These data are called RAW DATA. Number of leave 012345 Number of lecturers (frequency) 1294212 Data in the form of the frequency distribution table – ungrouped data

4. If the raw data have a large number of data – use grouped data Number of leave 0-12-34-5 Number of lecturers ( frequency) 2163 The frequency distribution above shows the same data but grouped into a few intervals

5. Terms 2. Class limit: - The lower limit is the smallest value of the class limit - The upper limit is the largest value of the class limit 1. Class interval: - Is the interval that includes all the value in a data set bounded by the lower and upper limit of the class

6. 3. Class Boundary: -Mid point of the upper limit of one class and the lower limit of the next class. Example: Lower boundary Upper boundary 1 - 2 3 - 45 - 6

7. 5. Class mark (middle value of a class): OR 4. Class width:

8. General method in constructing a frequency distribution table :- Step 1 : Decide the number of classes. Use Sturge’s Rule as a guideline. FREQUENCY DISTRIBUTION FOR GROUPED DATA A frequency distribution is a summary of how often each score occurs by grouping the score together Where k = number of class intervals n = the total number of observations in the data The number of k should be rounded up to the nearest integer

9. Step 4 : State the class limit for each class –I–It is chosen such that there is no overlapping between classes and each value can go to only one class. Step 3 : Find the class width The class width Step 2 : Find the value of range Range = Largest value – Lowest value.

10. Example 1: 3, 8, 1, 2, 2, 6, 7, 9, 1, 3 7, 9, 3, 4, 5, 6, 8, 2, 1, 4 Construct the frequency distribution table for the above data Solution : Step 1: Using Sturge’s rule, the number of class interval is 5 class intervals

11. Step 2 : Find the value of range. Range = Largest value – Lowest value = 9 – 1 = 8 The class width A suitable class width is 2 Step 3 : Find the class width

12. Frequency 6 5 3 4 2 Class intervalTally 1 - 2//// / 3 – 4//// 5 – 6/// 7 – 8//// 9 -10// The frequency distribution table 3, 8, 1, 2, 2, 6, 7, 9, 1, 3,7, 9, 3, 4, 5, 6, 8, 2, 1, 4 Step 4 : State the class limit for each class

13. Class intervalClass Boundary 1 - 20.5 – 2.5 3 – 42.5 – 4.5 5 – 64.5- 6.5 7 – 86.5 -8.5 9 -108.5 -10.5 Class Mark Frequency (f) 1.56 3.55 5.53 7.54 9.52 Class limits Lower limits Upper limits

14. RELATIVE FREQUENCY DISTRIBUTION - To compare the difference between two sets of data - The percentage of the frequency for each class compared to the total frequency Notes :

15. Class intervalFrequency 1 - 26 3 – 45 5 – 63 7 – 84 9 -102 Relative frequency

16. Below are the steps as a guide to construct a histogram (a).Construct a frequency distribution table and include a column of class boundary (b).Draw a horizontal line for the class boundaries and label it with the variable name (c).Draw a vertical line for the frequency (d).Draw the frequency in the form of vertical bar corresponding to each class boundary

17. Example 2 The waiting time for 50 patients who are seeking treatment from a doctor in a clinic is shown in the following frequency distribution: Waiting timeNumber of patients 5-93 10-142 15-1922 20-2418 25-294 30-341 Draw a histogram to represent the above information

18. Waiting time Number of patients 5-93 10-142 15-1922 20-2418 25-294 30-341 Solution: Class boundaries Class width 4.5-9.55 9.5-14.55 14.5-19.55 19.5-24.55 24.5-29.55 29.5-34.55 The height of the rectangles represent the frequency

19. Waiting time (minutes) Number of patient 5 10 15 20 4.59.514.519.524.529.534.5 Histogram for waiting time for 50 patients Class boundaries Frequency (f)

20. Frequency Polygon - Is obtain by connecting the midpoint ( or class mark) of each class at the top of the bar in the histogram Waiting time (minutes) Number of patient 5 10 15 20 4.59.514.519.524.529.534.5

21. Frequency Polygon Waiting time (minutes) Number of patient 5 10 15 20 4.59.514.519.524.529.534.5

22. Cumulative Frequency Curve ( Ogive) ~ Cumulative frequency is the sum of the frequencies accumulated up to the upper class boundary of each class. ~ The cumulative frequencies are plotted against the upper class boundaries. ~ The cumulative frequency is used to determine the number of observations which lie below a certain upper class boundary.

23. REMARKS…………………………… ~Ogive can be drawn based on : Class boundary versus cumulative frequency Class boundary versus percentage Class boundary versus relative frequency ~ There are various ways of writing the class boundary.

24. Waiting timeNumber of patients 5-93 10-142 15-1922 20-2418 25-294 30-341 Draw a ‘less than’ cumulative frequency curve based on the above information Example 3(a) :

25. Solution Waiting time Number of patients x<4.5- 4.5 - 9.53 9.5 - 14.52 14.5 - 19.522 19.5 - 24.518 24.5 - 29.54 29.5 - 34.5 1 Cumulative frequency 0303 5 27 45 49 50

26. * 45 40 35 30 25 20 15 10 5 0 Number of patients / cumulative frequency 9.5 14.5 19.5 24.5 29.5 34.5 Marks / Upper boundries * * * * * * 4.5 *

27. Solution Class BoundaryCumulative frequency percentage(%) x<4.500 X<9.536 x<14.5510 X<19.52754 X<24.54590 X<29.54998 X<34.550100 3(b) Draw a cumulative frequency percentage curve based on the above information

28. 100 90 80 70 60 50 40 30 20 10 0 cumulative frequency percentage 9.5 14.5 19.5 24.5 29.5 34.5 Marks / Upper boundries * * * * * * P 25 1619 23.4 Q3Q3 P 50 4.5 *

29. Solution Class BoundaryCumulative frequency Relative cumulative frequency x<4.500 4.5<x<9.530.06 9.5<x<14.550.10 14.5<x<19.5270.54 19.5<x<24.5450.90 24.5<x<29.5490.98 34.5<x<34.5501 3(c) Draw a relative cumulative frequency curve based on the above information

30. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Relative cumulative frequency 9.5 14.5 19.5 24.5 29.5 34.5 Marks / Upper boundries * * * * * * 4.5 *

31. Mass (kg)Frequency 39.5 – 44.53 44.5 – 49.52 49.5 – 54.57 54.5 – 59.518 59.5 – 64.518 64.5 – 69.5 69.5 – 74.5 3131 Example 4 The table below shows the frequency distribution of the mass of 52 students at a college

32. Draw a ‘less than’ cumulative frequency curved based on the above information. Estimate from the graph, (a)How many students were of mass less than 57 kg? (b)How many students were of mass more than 61 kg? (c)What was the mass exceeded by 20 % of the students?

33. Mass (kg)Frequency x < 39.5 x < 44.5 0303 x < 49.55 x < 54.512 x < 59.530 x < 64.548 x < 69.5 x < 74.5 51 52 Solution

34. Upper boundary The mass of 52 students at a college Number of students

35. (a)From the graph, there are 20 students whose mass less than 57 kg. (b) From the graph, there are 36 students whose mass less than 61kg. The total no. of students are 52, therefore the no. of students whose mass more than 61 kg are 52 – 36 = 16 (c) If 20% students exceeded the mass: Let x be the mass of the student: 0.2 X 52 = 10.4 = 10 students 52 – 10 = 42 Since the ogive is ‘less than’ we have 42 students whose mass less than x. From the graph, x = 63 kg.