It would be 3*5 = 15.
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
124
score of 92
A large standard deviation means that the data were spread out. It is relative whether or not you consider a standard deviation to be "large" or not, but a larger standard deviation always means that the data is more spread out than a smaller one. For example, if the mean was 60, and the standard deviation was 1, then this is a small standard deviation. The data is not spread out and a score of 74 or 43 would be highly unlikely, almost impossible. However, if the mean was 60 and the standard deviation was 20, then this would be a large standard deviation. The data is spread out more and a score of 74 or 43 wouldn't be odd or unusual at all.
1.50
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
The mean is 46.
Because the z-score table, which is heavily related to standard deviation, is only applicable to normal distributions.
It is the normalised Gaussian distribution. To speak of a 'standard z' distribution is somewhat redundant because a z-score is already standardised. A z-score follows a normal or Gaussian distribution with a mean of zero and a standard deviation of one. It's these specific parameters (this mean and standard deviation) that are considered 'standard'. Speaking of a z-score implies a standard normal distribution. This is important because the shape of the normal distribution remains the same no matter what the mean or standard deviation are. As a consequence, tables of probabilities and other kinds of data can be calculated for the standard normal and then used for other variations of the distribution.
If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.
z = (75 - 85)/5 = -10/5 = -2
It depends on the underlying distribution. If Gaussian (standrad normal) then the percentile is 77.
It is the value that is one standard deviation greater than the mean of a Normal (Gaussian) distribution.
z=(x-mean)/(standard deviation of population distribution/square root of sample size) T-score is for when you don't have pop. standard deviation and must use sample s.d. as a substitute. t=(x-mean)/(standard deviation of sampling distribution/square root of sample size)
124
The standardised score decreases.
IQ is distributed normally, with a mean of 100 and a standard deviation of 15. The z-score of 100 is therefore:(value-mean)*standard deviation= (100-100)*15= 0More generally, a raw score that is equivalent to the mean of a normal distribution will always have a z-score of 0.