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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)?

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2 Answers

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%
 
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
 
 

Comments

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.
I understand now the problem. Thank you very much for your responses and best to you.

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