The answer will depend on what the distribution is! And since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.
To see how wide spread the results are. If the average (mean) grade for a certain test is 60 percent and the standard deviation is 30, then about half of the students are not studying. But if the mean is 60 and the standard deviation is 5 then the teacher is doing something wrong.
mrs.sung gave a test in her trigonometry class. the scores were normally distributed with a mean of 85 and a standard deviation of 3. what percent would you expect to score between 88 and 91?
There is insufficient information in the question to answer it. In order to compute a mean and a standard deviation, you need at least two data points, but the question only gave one. Please restate the question.
to find percent deviation you divide the average deviation into the mean then multiply by 100% . to get the average deviation you must subtract the mean from a measured value.
No.
Percent variation is the standard deviation divided by the average
Percent deviation formula is very useful in determining how accurate the data collected by research really is. Percent Deviation = (student data-lab data) / lab data then multiplied by 100 Note: If the percent deviation is a negative number that means the student data is lower than the lab value.
To see how wide spread the results are. If the average (mean) grade for a certain test is 60 percent and the standard deviation is 30, then about half of the students are not studying. But if the mean is 60 and the standard deviation is 5 then the teacher is doing something wrong.
It is 68.3%
yes
s
20 percent
false
mrs.sung gave a test in her trigonometry class. the scores were normally distributed with a mean of 85 and a standard deviation of 3. what percent would you expect to score between 88 and 91?
67% as it's +/- one standard deviation from the mean
There is insufficient information in the question to answer it. In order to compute a mean and a standard deviation, you need at least two data points, but the question only gave one. Please restate the question.
to find percent deviation you divide the average deviation into the mean then multiply by 100% . to get the average deviation you must subtract the mean from a measured value.