Easy. The mean deviation about the mean, for any distribution, MUST be 0.
standard normal
It is any standardised distribution.
T-scores have a mean of 50 and a standard deviation of 10. These values are fixed and do not change regardless of the distribution of T-scores.
Step 1: Find the mean Step 2: Find the deviation from the mean Step 3: Take the absolute value of the deviation Step 4: Find the mean of the absolute deviation. x----x-mean 63 63-63 0 69 69-63 6 62 62-63 -1 57 57-63 -6 64 64-63 1 mean = (63+69+62+57+64)/5 = 63 Taking the absolute deviations, we have 0,6,1,6,1 Averaging these deviations : (0+6+1+6+1)/5 =14/5 = 2.8 Mean absolute deviation = 2.8
Mean 0, standard deviation 1.
Mean = 0 Standard Deviation = 1
Easy. The mean deviation about the mean, for any distribution, MUST be 0.
No.
standard normal
It is any standardised distribution.
Standard deviation doesn't have to be between 0 and 1.
a is true.
T-scores have a mean of 50 and a standard deviation of 10. These values are fixed and do not change regardless of the distribution of T-scores.
z=(x-mu)/s = (-10+9)/2 z = -1/2 Note that the standard normal has a mean of 0, therefore: The value of -10 is to the left of the mean of -9 The value of -1/2 is to the left of the mean of 0.
T scores are also standardized norm scores, where the mean value is 50 and standard deviation value is 10, in contrast to Z scores where mean value is "0" and standard deviation value is 1. -Rama Reddy Karri
standard deviation only measures the average deviation of the given variable from the mean whereas the coefficient of variation is = sd\mean Written as "cv" If cv>1 More variation If cv<1 and closer to 0 Less variation