Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, normal random v

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 22 bags?
(b)The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 22 bags is below 5.95 ounces?
(c)What should the mean fill volume equal in order that the probability that the average of 22 bags is below 6.1 ounces is 0.001?

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Tài Đức 4 years 2021-08-28T15:44:33+00:00 1 Answers 115 views 0

Answers ( )

  1. Thunguyet
    0
    2021-08-28T15:46:02+00:00
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    Answer:

    a) 0.0171 fluid ounces.

    b) 0% probability that the average fill volume of 22 bags is below 5.95 ounces

    c) The mean should be of 6.153 fluid ounces.

    Step-by-step explanation:

    To solve this question, we need to understand the normal probability distribution and the central limit theorem.

    Normal Probability Distribution:

    Problems of normal distributions can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

    Central Limit Theorem

    The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    Standard deviation of 0.08 fluid ounces.

    This means that \sigma = 0.08

    (a)What is the standard deviation of the average fill volume of 22 bags?

    This is s when n = 22. So

    s = \frac{\sigma}{\sqrt{n}}

    s = \frac{0.08}{\sqrt{22}}

    s = 0.0171

    (b)The mean fill volume of the machine is 6.16 ounces, what is the probability that the average fill volume of 22 bags is below 5.95 ounces?

    We have that \mu = 6.16. The probability is the p-value of Z when X = 5.95. So

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{5.95 - 6.16}{0.0171}

    Z = -12.3

    Z = -12.3 has a p-value of 0.

    0% probability that the average fill volume of 22 bags is below 5.95 ounces.

    (c)What should the mean fill volume equal in order that the probability that the average of 22 bags is below 6.1 ounces is 0.001?

    X = 6.1 should mean that Z has a p-value of 0.001, so Z = -3.09. Thus

    Z = \frac{X - \mu}{s}

    -3.09 = \frac{6.1 - \mu}{0.0171}

    6.1 - \mu = -3.09*0.0171

    \mu = 6.153

    The mean should be of 6.153 fluid ounces.

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