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# Using statistics, find mean and probabilities

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)?

### 2 Answers by Expert Tutors

Ryan S. | Mathematics and StatisticsMathematics and Statistics
4.8 4.8 (10 lesson ratings) (10)
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%

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (628 lesson ratings) (628)
1
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