Question

5.4.37 EQuestion Help Your lumber company has bought a machine that automatically cuts lumber. The seller of the machine claims that the machine cuts lumber to a mean length of 8 feet (96 inches) with a standard deviation of 0.6 inch. Assume the lengths are normally distributed. You randomly select 42 boards and find that the mean length is 96.17 inches. Complete parts (a) through (C) EE Click the icon to view page 1 of the standard normal table EE Click the icon to view page 2 of the standard normal table (a) Assuming the seller's claim is correct, what is the probability that the mean of the sample is 96.17 inches or more? (Round to four decimal places as needed)

Answer 1

5.4.37a)

µ =    96                          
σ =    0.6                          
n=   42                          
right tailed                              
X ≥   96.17                          
                              
Z =   (X - µ )/(σ/√n) =   1.836                      
                              
P(X ≥   96.17   ) = P(Z ≥   1.836   ) =   P ( Z <   -1.836   ) =    0.0332

question- 5.4.38

µ =    47000              
σ =    800              
n=   100              
left tailed                  
X ≤    46852              
                  
Z =   (X - µ )/(σ/√n) =   -1.850          
                  
P(X ≤   46852   ) = P(Z ≤   -1.850   ) =   0.0322