PDE110 ASSIGNMENT
NAME OF STUDENT: EME IHUOMA
Admission number :NTI/PGDE/2016/1001
Questions
1) Arrange the numbers 12,56,42,21,5,18.10,3,61,34,65,24 in an array and 2) Determine the range * If the class marks in a frequency distribution of length of laurels are 129 138 147 156 165 174 and 183 mm, find the class interval size boundries and limits
Solutions
* 65
61
56
42
34
24
21
18
12
10
5
3
* Range = H-L WHERE H =Highest value L= Lowest value Therefore; Range = 65-3=62 Range =62

Solutions (continues) A Length of laurel mm | | 129 | | 138 | | 147 | | 156 | | 165 | | 174 | | 183 | | * Class Interval size is the difference between the upper class and the lower class. From the data we have been given above the true value end points are 128.5-137.5. Making 128.5 the lower class and 137.5 the upper class. Therefore; Class Interval = 137.5-128.5= 9 Class Interval=9 * Class limits are 129 -137, 138- 146, 147-155, 156-164, 165-173, 174-182, 183 * 137+138
2
137+138
2
Class boundary is the sum of the upper class limit of one class to the lower class limit of the next higher class. And then dividing the sum by two. The first class upper class limit is 137 and second class lower class limit is 138. Therefore; class boundaries = = 137.5 146+147
2
146+147
2
The second class upper class limit is 146 and the third class lower class limit is 147. Therefore; class boundary= = 146.5

The third class upper class limit is 155 and the fourth class lower limit is 156 155+156
2
155+156
2