Mean and Standard Deviation
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Free InquiryAn experiment was conducted to test the quality of a plant seed in an agricultural farm. Below data shows the number of survived plants from a sample of 50 equally designed seed beds. The principal researcher wants to prepare a quick presentation about the outcome of the experiment.
20 |
15 |
22 |
3 |
13 |
27 |
11 |
12 |
11 |
14 |
12 |
4 |
16 |
4 |
15 |
2 |
22 |
14 |
1 |
13 |
10 |
8 |
23 |
17 |
7 |
21 |
22 |
26 |
21 |
4 |
23 |
14 |
21 |
22 |
18 |
3 |
7 |
6 |
12 |
5 |
13 |
21 |
6 |
23 |
12 |
2 |
8 |
16 |
6 |
4 |
a) Construct a Frequency Distribution Table with a class interval of 5 for the survived plants among those 50 seed beds using the following format. (5 marks)
Class Interval |
Mid values Marks |
Frequency |
0 to less than 5 |
||
5 to less than 10 |
||
10 to less than 15 |
||
15 to less than 20 |
||
20 to less than 25 |
b) Compute Mean and Standard Deviation from the above table. (4 marks)
Hints: Mean = ,
Standard Deviation =
Where n = number of values or sample size
c = number of classes in the frequency distribution
mj = midpoint of the jth class
fj = number of values in the jth class]
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