Question

ne the class interval (CI), the width of classes of it may be the difference between any wo succeeding class midpoints (CM). The CM is the average between the limits of a class. For on, CI must be a whole number. It can be obtained by the equation (3) Determi each computati CI-R/K-101.5/10-10.15 (4) Take the frequency distribution for each set of data. On the basis of this method, the frequency distribution is as follows: Table 8.1. Frequency Distribution Frequency 41-51 52-62 13 3 17 63-73 74-84 85-95 96-106 07-117_ 118- 128 129-139 140-150 12 15 48 63 3 6467 43 49 17 26 5 10 18 20 20 198 256 Total (5) Construct a frequency polygon and a histogram for each set of data on the same graphing paper. Assign class midpoints on the abscissa and frequencies on the ordinates. A histogram is a series of vertical rectangle magnitude vary according to the frequency of each class, while a frequency polygon is a smooth curve made by joining the midpoints of the upper horizontal sides of the rectangles in a histogram F2 FI P2 Legend: PI 100% 90% T 60% 30% 90 57 68 79 90 101 112 123 134 145 Ps, Fi and Fa data Figure 8.1: Frequeney polygon and histogram of P

Answer 1

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