This question was reported to be well attempted by majority the candidates who were said to
have demonstrated a good knowledge of the topic. Majority of the candidates were able to
obtain the class mark for each interval, calculate the mean height and find the required
probability.
Heicht |
Class Mark (x) |
Freauencv (f) |
fx |
1.40 -1-42 |
1.41 |
2 |
2.82 |
1.43 -1.45 |
1.44 |
4 |
5.76 |
1.46 - 1.48 |
1.47 |
19 |
27.93 |
1.49-1.51 |
1.50 |
30 |
45.00 |
1.52 -1.54 |
1.53 |
24 |
36.72 |
1.55 -1.57 |
1.56 |
14 |
21.84 |
1.58 -1.60 |
1.59 |
6 |
9.54 |
1.61 -1.63 |
1.62 |
1 |
1.62 |
|
|
>. f = 100 |
L Ix = 151.23 |
Mean = ∑fx/∑f = 151.23 = 1.5123
100
Probability (height> mean height) = 45/100 = 0.45 or 9/20
