Question

The amount of soft drink that goes into a typical 355 ml can varies from can tocan. It is normally distributed with an adjustable mean μ and a fixed standard deviation of 0.23 ml. (The adjustment is made to the filling machine.)

a)If the regulations require that 96.93% of the cans have at least 355ml, what is the smallest mean μ that can be used to meet the regulations (to 2 decimal places)? Show logic / equations / calculations (including values on diagram)

b)If the mean setting from part a) is used, what is the likelihood (in percentage) that a typical can will have more than 355.8 ml (to 2 decimal places)? Show logic/equations/calculations

Answer 1

a) P(X≥355) = 0.9693

P(X<x) = 1 - 0.9693 = 0.0307

z value at 0.0307= -1.8706     (excel formula =NORMSINV(0.0307))

µ=X-Zσ = 355+1.8706*0.23 =  355.43

b)

µ =    355.43   
σ =    0.23                  
                      
P ( X ≥   355.8   ) = P( (X-µ)/σ ≥ (355.8-355.43) / 0.23)              
= P(Z ≥   1.61   ) = P( Z <   -1.608   ) =    0.0540   (answer)