P198 Intravenous fluid bags are filled by an automated filli

P.198

Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, normal random variables with a standard deviation of 0.08 fluid ounces.

(a)What is the standard deviation of the average fill volume of 21 bags?
(b)The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 21 bags is below 5.95 ounces?
(c)What should the mean fill volume equal in order that the probability that the average of 21 bags is below 6.2 ounces is 0.001?

Solution

a)

standard error of the mean = s/ sqrt(n) = 0.08/sqrt(21) = 0.017457431 [answer]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    5.95      
u = mean =    6.16      
n = sample size =    21      
s = standard deviation =    0.08      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -12.0292612      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -12.0292612   ) =    1.24691*10^-33 or VERY CLOSE TO 0. [ANSWER]

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c)

For a left tailed area of 0.001, the critical z score is

zcrit = -3.090232306

Thus, as

z = (X - u) sqrt(n) / s

Then

u = X - z s / sqrt(n)

= 6.2 - (-3.090232306) (0.08) / sqrt(21)

= 6.253947518 [ANSWER]