z = - 0.8416 to z = + 0.8416
false
In a normal distribution half (50%) of the distribution falls below (to the left of) the mean.
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 standard normal distribution is a special case of the normal distribution. The standard normal has mean 0 and variance 1.
The domain of the normal distribution is infinite.
false
-0.772 < Z < 0.772
In a normal distribution half (50%) of the distribution falls below (to the left of) the mean.
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...)
In the normal distribution, the mean and median coincide, and 50% of the data are below the mean.
No. By definition of the median, the median has 50 percent of the case below and 50 percent of the cases above. This has nothing to do with the cases being in a normal distribution.
A normal distribution is symmetrical; the mean, median and mode are all the same, on the line of symmetry (middle) of the graph.
-1.28
A normal distribution is symmetric and when looked at on a graph, the graph looks like a bell shaped curve. Approximately 95 percent of its values should lie within two standard deviations of the mean. Frequency of the data lies mostly in the middle of the curve.
The standard normal table tells us the area under a normal curve to the left of a number z. The tables usually give only the positive value since one can use symmetry to find the corresponding negative values. The middle 60 percent leaves 20 percent on either side. So we want the z scores that correspond to that 80 percentile which is .804. Therefore the values are are between z scores of -.804 and .804 * * * * * I make it -0.8416 to 0.8416
Divergent Boundaries are where the data doesn't fit the normal distribution. The point at which the data diverges is the divergent Boundary.
It is 68.3%