Using statistics, find mean and probabilities

Question

Soft Drink Cans
 
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)?

Answer 1

We want P(x>=355)=.9925 which implies P(x<=355)=1-.9925=.0075

 

This corresponds to the standard normal distribution where Φ(z)=.0075.

 

Using a table, calculator, or computer we find that z=-2.43

 

z=(x-μ)/σ

-2.43 = (355-μ)/0.4

 

Solving for μ give μ = 355.972 ml

 

Now we want for x~N(355.972,0.4) P(x>=357) = 1-P(x<=357)

 

z=(357-355.972)/0.4=2.57

 

Using a table, calculator, or computer we find Φ(z)=0.9949

 

So 1-Φ(z)=0.0051=0.51%

 

Answer 2

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. 

Since z = (355- mean)/.4

We get mean = 354.0270484