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 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.
An average IQ score for a 7 year old is typically around 100, so a score of 119 would be considered above average. IQ scores are standardized to have a mean of 100 with a standard deviation of 15, so a score of 119 would fall about 1 standard deviation above the average.
score of 92
The absolute value of the standard score becomes smaller.
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.
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.
standard deviation
There are approximately 16.4% of students who score below 66 on the exam.
2 standard deviation's below the mean
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.