You can find this using a table of z scores.
Z = (x - mu) / s
Z = (109 - 100) /15 = 9/15 = 3/5 = .6
You want the percentage of students having a z score above .6.
p = 1 - .7257 = .2743
.2743 * 1800 = 493 students with a score above 109
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
The absolute value of the standard score becomes smaller.
2 standard deviation's below the mean
Because the z-score table, which is heavily related to standard deviation, is only applicable to normal distributions.
78
.The test has a mean, or average, standard score of 100 and a standard deviation of 16 (subtests have a mean of 50 and a standard deviation of 8). The standard deviation indicates how far above or below the norm the subject's score is.
The standardised score decreases.
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
score of 92
A z-score cannot help calculate standard deviation. In fact the very point of z-scores is to remove any contribution from the mean or standard deviation.
The absolute value of the standard score becomes smaller.
When you don't have the population standard deviation, but do have the sample standard deviation. The Z score will be better to do as long as it is possible to do it.
A negative Z-Score corresponds to a negative standard deviation, i.e. an observation that is less than the mean, when the standard deviation is normalized so that the standard deviation is zero when the mean is zero.
There are approximately 16.4% of students who score below 66 on the exam.
standard deviation
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
2 standard deviation's below the mean