Determine which expression(s) is in its simplified form. Select all situations that apply.
2 answers:
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Answer:
Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:
Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:
Class 66 - 68:
When the computation is completed, the frequency distribution will be:
Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:
<em>See attachment for result of 1-VarStats</em>
