Question

A soda machine is regulated to dispense an avarage of 240ml with an standar deviation of 15 mL. The machibe is verified periodically taking samples of 40 drinks to wich the mean is calculated. If the mean of the sample value is in the u(-+)2(sigma/sqrt(n)) the machine is estimated to be operating correctly. If the avarage of the sample is out of the interval the machine need to be ajusted.

A.Calculate the probability of the machine needing to be ajusted.

B. If the avarge of a sample is of 236 ml what is needed to be done?

C. Which is the needed interval for only 1% of the times the machine requires ajustment

Answer 1

A)

µ =    240                              
σ =    15                              

If the mean of the sample value is in the u(-+)2(sigma/sqrt(n)) the machine is estimated to be operating correctly

µ± z σ

so, z =±2/√(n) = ±2/√40 = 0.316

   P (    -0.316 < Z <    0.316   )
                                  
= P ( Z <    0.316   ) - P ( Z <   -0.316   ) =    0.62409   -    0.375915   =    0.2482

so, P(machine needing to be ajusted) = 1 - 0.2482 = 0.7518

B)

If the mean of the sample value is in the u(-+)2(sigma/sqrt(n)) the machine is estimated to be operating correctly

µ±2*σ/√n = (235.26 ,244.74)

since, 236 falls in interval,

so, the machine is estimated to be operating correctly

C)

µ =    240                          
σ/√n =    2.372                          
proportion=   0.01  

                      
proportion left    0.99   is equally distributed both left and right side of normal curve                       
z value at   0.99/2 = ±   0.0125   (excel formula =NORMSINV(   0.99   / 2 ) )      
               
z = ( x - µ ) / σ/√n   
so, X = z σ/√n + µ =

                              
X1 =   -0.0125   *   2.372   +   240   =   239.97  
X2 =   0.0125   *   2.372   +   240   =   240.03  

so, interval is (239.97 , 240.03)