First, 0.01% is not in the fourth sd.
Prob(-4 < Z < 4) is approximately 0.0064 or 0.64% which, when rounded to 0.1% but is, in fact, smaller.
Prob(0 < Z < 5) is approximately 0.000 000 287 = 0.000 028 7%
50 percent
The answer is about 16% Using the z-score formula(z = (x-u)/sd) the z score is 1. This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
(x-400)/100=1.882 x=588.2
A normal distribution with a mean of 65 and a standard deviation of 2.5 would have 95% of the population being between 60 and 70, i.e. +/- two standard deviations.
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.
It is 68.3%
50 percent
false
For data sets having a normal distribution, the following properties depend on the mean and the standard deviation. This is known as the Empirical rule. About 68% of all values fall within 1 standard deviation of the mean About 95% of all values fall within 2 standard deviation of the mean About 99.7% of all values fall within 3 standard deviation of the mean. So given any value and given the mean and standard deviation, one can say right away where that value is compared to 60, 95 and 99 percent of the other values. The mean of the any distribution is a measure of centrality, but in case of the normal distribution, it is equal to the mode and median of the distribtion. The standard deviation is a measure of data dispersion or variability. In the case of the normal distribution, the mean and the standard deviation are the two parameters of the distribution, therefore they completely define the distribution. See: http://en.wikipedia.org/wiki/Normal_distribution
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
The answer is about 16% Using the z-score formula(z = (x-u)/sd) the z score is 1. This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
Percent variation is the standard deviation divided by the average
Not possible to tell you without knowing how many students' there are, and what distribution you wish to use (i.e normal distribution, t-distribution etc...)
The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%
(x-400)/100=1.882 x=588.2
A normal distribution with a mean of 65 and a standard deviation of 2.5 would have 95% of the population being between 60 and 70, i.e. +/- two standard deviations.
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.