An operation manager at an electronics company wants to test

An operation manager at an electronics company wants to test their ampifiers. The design engineer claims they have a mean output of 482 wants with variance of 121. What is the probability that the mean amplifier output would differ from the population mean by greater than 3.2 watts in a sample of 59 amplifiers if the claim is true? Round your answer to your decimal places.

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 482-3.2 =   478.8      
x2 = upper bound = 482+3.2 =   485.2      
u = mean =    482      
n = sample size =    59      
s = standard deviation =    11      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.234515127      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.234515127      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.012724599      
P(z < z2) =    0.987275401      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.974550802   [ANSWER]