SOLUTIONS TO FORM FOUR SERIES ONE
WEEKLY QUESTIONS:
QN 1:;
solution
(a) Three significant figures 300.0
(b) Three decimal places 300.326
QN2:
solution
Rationalization of the denominator
QN3:
(a) solution
(b) solution
QN4 (a) solution
Table of values for the function
(b) The domain and ranges for the
relation are such that
QN5.
Solution:
Marks | Class Marks(x) | Frequency (f) | f.x | Cumulative Frequency |
0-9 | 4.5 | 1 | 4.5 | 1 |
10-19 | 14.5 | 2 | 29 | 3 |
20-29 | 24.5 | 5 | 122.5 | 8 |
30-39 | 34.5 | 11 | 379.5 | 19 |
40-49 | 44.5 | 21 | 934.5 | 40 |
50-59 | 54.5 | 20 | 1090 | 60 |
60-69 | 64.5 | 17 | 1096.5 | 77 |
70-79 | 74.5 | 10 | 745 | 87 |
80-89 | 84.5 | 6 | 507 | 93 |
90-99 | 94.5 | 4 | 378 | 97 |
100-109 | 104.5 | 4 | 418 | 101 |
110-119 | 114.5 | 1 | 114.5 | 102 |
TOTAL | 102 | 5819 |
(a)Histogram and Frequency Polygon see
figure 1.
(b) The modal class refers to the class
with the highest frequency which in this case is 40-49.
(c) Solution:
(d) Calculating the median
But before anything you should find the
median class by calculating the middle term of the cumulative
frequency which is usually the number which is equal to the half of
the total of frequency or just above.
Which in this case 102/2 = 51
So we choose the class with a
cumulative frequency of 60.
Hence the median class is 50-59
Histogram and Frequency Polygon
FREQUENCY POLYGON
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Chapisha Maoni