You measure a velocity of some particles many thousands of t

You measure a velocity of some particles many thousands of times, and find that the average is 90.7 m/s, with a standard deviation of 2.0 m/s. Now, you make an additional 20 measurements, and take the average just of those 20 measurements. What is the probability that you find that average to be either greater than 91.6 m/s, or less than 89.8 m/s? Then consider the ensemble of the 20 additional measurement like a single measurement yet with a reduced (how large?) uncertainty.

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    89.8      
x2 = upper bound =    91.6      
u = mean =    90.7      
n = sample size =    20      
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.01246118      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.01246118      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022085672      
P(z < z2) =    0.977914328      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.955828655      

Thus, those outside this interval is the complement = 0.044171345   [ANSWER]