A company produces steel rods The lengths of the steel rods

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 145-cm and a standard deviation of 1.3-cm. For shipment, 22 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 144.7-cm and 145.4-cm.

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    144.7      
x2 = upper bound =    145.4      
u = mean =    145      
n = sample size =    22      
s = standard deviation =    1.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.082403637      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.443204849      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.139536606      
P(z < z2) =    0.925518613      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.785982007   [ANSWER]