Intravenous fluid bags are filled by an automated filling ma

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 20 bags?
(b)The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 20 bags is below 5.95 ounces?
(c)What should the mean fill volume equal in order that the probability that the average of 20 bags is below 6.2 ounces is 0.001?

Solution

Intravenous fluid bags are filled by an automated filling machine.

Let X be an random variable.

X : Fill volumes of the bags.

X is independent and normal random variable with standard deviation is 0.08 ounces.

n = Number of bags = 20.

standard deviation = 0.08 ounces.

a) What is the standard deviation of the average fill volume of 20 bags?

Earlier we define that X is fill volumes of the bags.

And we have given that average fill volume of 20 bags i.e. Xbar.

W know that the result Xbar ~ N ( µ , ² / n )

Mean of Xbar = µ

and variance of Xbar = ² / n

standard deviation = / sqrt(n) (Where sqrt(n) means square root of n)

Standard deviation = 0.08 / sqrt(20)

= 0.02.

b) The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 20 bags is below 5.95 ounces?

It means we have given that µ = 6.16 ounces.We have to find the P(Xbar < 5.95).

P(Xbar < 5.95 ) = P( (Xbar - µ) / ( / sqrt(n)) < (5.95 - µ) / ( / sqrt(n))

= P( Z < (5.95 - 6.16) /(0.02)) _____(From a) part).

= P( Z < -10.5 )

This probability we can calculate using excel which has command NORMSDIST ( z = -10.5) which guves us probability 4.31901 * 10^-26 .which is approximately equal to 0.

Therefore P(Xbar < 5.95 ) = 0.

c) What should the mean fill volume equal in order that the probability that the average of 20 bags is below 6.2 ounces is 0.001?

Here we have to find mean fill volume i.e. µ.

And we have given that P(Xbar < 6.2) = 0.001

It means we have to find µ from given probability.

P(Xbar < 6.2) = P( (Xbar - µ) / ( / sqrt(n)) < (6.2 - µ) / ( / sqrt(n)) )

= P( Z < (6.2 - µ) / 0.02 ) _____(From part a)).

And we have given that P( Z < (6.2 - µ) / 0.02 ) =0.001.

It means for given values of Xbar , / sqrt(n) and probability we have to find the mean.

Here we use inverse normal probability which gives us z.

Excel command for this is NORMINV(probablility = 0.001 , mean = 6.2 , standard deviation = 0.08)

Which gives us z is 5.952781.

That is P( Z < (6.2 - µ) / 0.02 ) = 5.952781.

Compare z with (6.2 - µ) / 0.02.

(6.2 - µ) / 0.02 = 5.952781.

6.2 - µ = 5.952781 * 0.02

6.2 - µ = 0.1191

µ = 6.2 - 0.1190 = 6.081.

It means when 6.081 is the mean fill volume given that the probability that the average of 20 bags is below 6.2 ounces is 0.001.