In a learning experiment untrained mice are placed in a maze

In a learning experiment, untrained mice are placed in a maze and the time required for each mouse to exit the maze is recorded. The average time for untrained mice to exit the maze is = 50 seconds and the standard deviation of their times is = 16 seconds. 64 randomly selected untrained mice are placed in the maze and the time necessary to exit the maze is recorded for each. What is the probability that the sample mean differs from the population mean by more than 3?

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          

x1 = lower bound = 50 - 3 =   47      
x2 = upper bound = 50 + 3 =   53      
u = mean =    50      
n = sample size =    64      
s = standard deviation =    16      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.5      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597      

Thus, those outside this interval is the complement =    0.133614403   [ANSWER]