the diameter of a ping pong ball is approximatley normally d

the diameter of a ping pong ball is approximatley normally distributed   with a mean of 1.32 and a standard deviation of .08   a random sample of 4 ping pong balls is selected

what is probablilty of sample mean being less that 1.27?

what is prob of sample mean being between 1.27 and 1.29?

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    1.27      
u = mean =    1.32      
n = sample size =    4      
s = standard deviation =    0.08      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.25      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.25   ) =    0.105649774 [answer]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    1.27      
x2 = upper bound =    1.29      
u = mean =    1.32      
n = sample size =    4      
s = standard deviation =    0.08      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.25      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    -0.75      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105649774      
P(z < z2) =    0.226627352      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.120977579   [answer]