The amount of soft drink that goes into a typical 355 ml can varies from can to
can. It is normally distributed with an adjustable mean µ and a fixed standard deviation of 0.4 ml. (The adjustment is made to the filling machine.)

a) If the regulations require that 99.25% of the cans have at least 355
ml, what is the smallest mean µ that can be used to meet the regulations (to 2
decimal places)?

b) If the mean setting from part a) is used, what is the probability that a typical
can will have at least 357 ml (to 2 decimal places)?

This problem is easily solved with a TI-84 graphing calculator. You want the invnorm() function which can be accessed from the catalog key (2nd zero). This function returns the z value needed to get the desired

integral of the normal distribution.

So invnorm(.9925) returns 2.43237906 which is the desired z score.

Intuitively, if we want a high probability that x>355 we need a mean higher than 355. The z you got was for P(x<=355) we want P(x>=355) you got the right value for z in your solution but the wrong sign.

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