A machine is set to put 5,000 feet of thread on spools. A standard deviation of

Question

A machine is set to put 5,000 feet of thread on spools. A standard deviation of 250 feet is allowed. To make sure the machine is working properly, it is checked each week by measuring the lengths of thread put on 64 randomly selected spools. If the mean of the 64 lengths is within 50 feet of 5,000, the machine is said to be working properly. Assuming the machine is NOT working properly. AND IS OUT OF ADJUSTMENT SO THAT IT NOW IT IS PUTTING ON AN AVERAGE OF 4925 (the true population meant FEET OF THREAD, what percent of the times it is checked will it fail to pass inspection? 5.48% 10.96% 78.8% .0003%

Explanation / Answer

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    4950      
x2 = upper bound =    5050      
u = mean =    4925      
n = sample size =    64      
s = standard deviation =    250      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0.8      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.788144601      
P(z < z2) =    0.999968329      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.211823727  
      
Thus, those outside this interval is the complement = 0.788176273 or 78.8% [ANSWER, OPTION 3]

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