Let X be the mean of a random sample of size n 48 from the

Solution

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    2      
          
Thus, the mean, variance, and standard deviations are          
          
u = mean = (b + a)/2 =    1      
s^2 = variance = (b -a)^2 / 12 =    0.333333333      
s = standard deviation = sqrt(s^2) =    0.577350269      
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    0.9      
x2 = upper bound =    1.1      
u = mean =    1      
n = sample size =    48      
s = standard deviation =    0.577350269      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.2      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11506967      
P(z < z2) =    0.88493033      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.76986066   [ANSWER]