3.5
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The mean deviation of ANY set of numbers must always be zero.
Easy. The mean deviation about the mean, for any distribution, MUST be 0.
It is any standardised distribution.
standard normal
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].
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.
It is any standardised distribution.
standard normal
a is true.
Standard deviation doesn't have to be between 0 and 1.
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].
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
5.142857143 is the mean.12.43956044 is the variance.3.526976104 is the standard deviation.