Question

(3) Determine the class interval (Cl), the width of classes of it may be the difference two succeeding class midpoints (CM). The CM is the average between the limits of a class. For each computation, CI must be a whole number. It can be obtained by the equation 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 Class Total P PFF 41 51 52-62 13 17 63-73 74-84 85-95 96-106 107-117 118-128 129-139 140 150 | 15 | 48 | 63 3 64 67 43 49 17 26 10 6 Total 18 20 20 198 256 (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 F1 Legend:-PI 100% 90% 80% 70% 50% 20% 10% 0% 90 57 68 79 101 112 123 134 145 Figure &.1: Frequency polygon and histogram of P, Ps, Fi and F data

Answer 1

(3) Class Interval is the difference between any two succeeding class midpoints = 57 – 46 = 11

(5) I have constructed the frequency polygon and histogram for each set of data. You may draw the same graph. It would look better if you match the graph line of each set with the corresponding vertical bars.

Histogram & Frequency Polygon 70 60 30 20 10 46 68 79 90 101 112 134 145 P1 P2 . ' P3 .P4 P1 Frequency P2 Frequency --P3 Frequency -P4 Frequency .