Question

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 23 bags? Round your answer to four decimal places. σx = (b) The mean fill volume of the machine is 6.16 ounces. What is the probability that the average fill volume of 23 bags is below 5.95 ounces? Round your answer to two decimal places. (c) What should the mean fill volume equal in order that the probability that the average of 23 bags is below 6.0 ounces is 0.001? Round your answer to two decimal places.

Answer 1

a) 0.0167

b) 0

c) 5.948

Explanation:

We are given the following information in the question:

Mean, μ = 6.16 ounces

Standard Deviation, σ = 0.08 ounces

We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) Standard deviation of 23 bags

`\displaystyle(S.D)/(√(23)) = (0.08)/(√(23)) = 0.0167`

b) P( fill volume of 23 bags is below 5.95 ounces)

P(x < 5.95)

`P( x < 5.96) = P( z < \displaystyle(5.95 - 6.16)/(0.0167)) = P(z < -12.57)`

`= 1 - P(z < 12.57)`

Calculation the value from standard normal z table, we have,

`P(x < 5.95) = 1 - 1 = 0`

c) P( fill volume of 23 bags is below 6 ounces) = 0.001

P(x < 6) = 0.001

`P( x < 6) = P( z < \displaystyle(6 - \mu)/(0.0167))`

Calculation the value from standard normal z table, we have,

`P( z \leq -3.09) = 0.001`

`\displaystyle(6 - \mu)/(0.0167) = -3.09\\\\\mu = 6 + (0.0167* -3.09) = 5.948`

If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.