A doctor randomly selected 90 out of 300 patients to take a fitness test. The mean grade of the fitness test was 43.7 with a standard deviation of 0.52. Find the margin of error for the mean fitness grade. Use the formula below where n is the sample size and s is the standard deviation, then round the answer to the nearest tenth. Margin of Error = 2(sn√)
Question
A doctor randomly selected 90 out of 300 patients to take a fitness test. The mean grade of the fitness test was 43.7 with a standard deviation of 0.52. Find the margin of error for the mean fitness grade. Use the formula below where n is the sample size and s is the standard deviation, then round the answer to the nearest tenth. Margin of Error = 2(sn√)
Solution
To find the margin of error for the mean fitness grade, we will use the given formula: Margin of Error = 2(s√n).
Here, s is the standard deviation and n is the sample size.
Given in the problem, s = 0.52 and n = 90.
Let's plug these values into the formula:
Margin of Error = 2 * 0.52 * √90
First, calculate the square root of 90, which is approximately 9.49.
Then, multiply this by 2 and 0.52:
Margin of Error = 2 * 0.52 * 9.49 = 9.86
Rounding this to the nearest tenth, we get 9.9.
So, the margin of error for the mean fitness grade is 9.9.
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