if standard deviation is 4 minutes 95% probability is about 2 standard deviations (actually 1.96) so you would need to allow 30 + 8 = 38 minutes
Assuming the returns are nomally distributed, the probability is 0.1575.
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.
Assuming the returns are nomally distributed, the probability is 0.1575.
30 percent.
in a normal distribution, the mean plus or minus one standard deviation covers 68.2% of the data. If you use two standard deviations, then you will cover approx. 95.5%, and three will earn you 99.7% coverage
Percent variation is the standard deviation divided by the average
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