a ball is thrown into the air. After t seconds, the ball's height is given by h(t)+=12-3(t-2) squared where h is measured in fee
t. What is the maximum height of the ball?
2 answers:
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<h3>Answer: 0.061</h3>
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Explanation:
Add up the values to get
1.96+1.81+1.97+1.83+1.87+1.84+1.85+1.94+1.96+1.81+1.86+1.95+1.89= 24.54
Then divide over 13 (the number of values) to get 24.54/13 = 1.8876923 which is approximate.
So the mean is approximately 1.8876923
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Now make a spreadsheet as shown below
We have the first column as the x values, which are the original numbers your teacher provided. The second column is of the form (x-M)^2, where M is the mean we computed earlier. We subtract off the mean and square the result.
After we compute that column of (x-M)^2 values, we add them up to get what is shown in the highlighted yellow cell at the bottom of the column.
That sum is approximately 0.04403076924
Next, we divide that over n-1 = 13-1 = 12
0.04403076924 /12 = 0.00366923077
That is the sample variance. Apply the square root to this to get the sample standard deviation. This is the point estimate of the population standard deviation. As the name implies, it works for samples that estimate population parameters.
sqrt(0.00366923077) = 0.06057417576822
This rounds to 0.061 which is the final answer.