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A bottling machine can be regulated so that it discharges an average of µu ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with o = 1.0 ounce. A sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each. Find the probability that the sample mean will be within .3 ounce of the true mean u for the chosen machine setting. If Y1, Y2, ..., Y9 denote the ounces of fill to be observed, then we know that the Y;'s are normally distributed with mean µ and variance o? = 1 for i = 1, 2, ...,9. Therefore, by Theorem 7.1, Y possesses a normal sampling distribution with mean µy = µ and variance o = o?/n = 1/9. We want to find P(Y – Hl < .3) = P[-,3 < (Y – µ) < .3] .3 Y - µ .3 = P| Because (Y – uy)/op = (Y – µ)/(a//n) has a standard normal distribution, it fol- lows that .3 .3 P(Y – Hl < .3) = P( 1//9 = P(-9 < Z <.9). Using Table 4, Appendix 3, we find How is it 1-2P(Z>0.9)? P(-9 .9) = 1 – 2(.1841) = .6318| Thus, the probability is only .6318 that the sample mean will be within .3 ounce of the true population mean.

A bottling machine can be regulated so that it discharges an average of µu ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with o = 1.0 ounce. A sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each. Find the probability that the sample mean will be within .3 ounce of the true mean u for the chosen machine setting. If Y1, Y2, ..., Y9 denote the ounces of fill to be observed, then we know that the Y;'s are normally distributed with mean µ and variance o? = 1 for i = 1, 2, ...,9. Therefore, by Theorem 7.1, Y possesses a normal sampling distribution with mean µy = µ and variance o = o?/n = 1/9. We want to find P(Y – Hl < .3) = P[-,3 < (Y – µ) < .3] .3 Y - µ .3 = P| Because (Y – uy)/op = (Y – µ)/(a//n) has a standard normal distribution, it fol- lows that .3 .3 P(Y – Hl < .3) = P( 1//9 = P(-9 < Z <.9). Using Table 4, Appendix 3, we find How is it 1-2P(Z>0.9)? P(-9 .9) = 1 – 2(.1841) = .6318| Thus, the probability is only .6318 that the sample mean will be within .3 ounce of the true population mean.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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This problem is related to the central limit theorem and all the steps and formulae are clear but for the last line with a 1-2P(Z>0.9) - how is this arrived at. Attached problem.

A bottling machine can be regulated so that it discharges an average of µu ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with o = 1.0 ounce. A sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each. Find the probability that the sample mean will be within .3 ounce of the true mean u for the chosen machine setting. If Y1, Y2, ..., Y9 denote the ounces of fill to be observed, then we know that the Y;'s are normally distributed with mean µ and variance o? = 1 for i = 1, 2, ...,9. Therefore, by Theorem 7.1, Y possesses a normal sampling distribution with mean µy = µ and variance o = o?/n = 1/9. We want to find P(Y – Hl < .3) = P[-,3 < (Y – µ) < .3] .3 Y - µ .3 = P| Because (Y – uy)/op = (Y – µ)/(a//n) has a standard normal distribution, it fol- lows that .3 .3 P(Y – Hl < .3) = P( 1//9 = P(-9 < Z <.9). Using Table 4, Appendix 3, we find How is it 1-2P(Z>0.9)? P(-9 <Z < 9) = 1– 2P(Z > .9) = 1 – 2(.1841) = .6318| Thus, the probability is only .6318 that the sample mean will be within .3 ounce of the true population mean.
Transcribed Image Text:A bottling machine can be regulated so that it discharges an average of µu ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with o = 1.0 ounce. A sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each. Find the probability that the sample mean will be within .3 ounce of the true mean u for the chosen machine setting. If Y1, Y2, ..., Y9 denote the ounces of fill to be observed, then we know that the Y;'s are normally distributed with mean µ and variance o? = 1 for i = 1, 2, ...,9. Therefore, by Theorem 7.1, Y possesses a normal sampling distribution with mean µy = µ and variance o = o?/n = 1/9. We want to find P(Y – Hl < .3) = P[-,3 < (Y – µ) < .3] .3 Y - µ .3 = P| Because (Y – uy)/op = (Y – µ)/(a//n) has a standard normal distribution, it fol- lows that .3 .3 P(Y – Hl < .3) = P( 1//9 = P(-9 < Z <.9). Using Table 4, Appendix 3, we find How is it 1-2P(Z>0.9)? P(-9 <Z < 9) = 1– 2P(Z > .9) = 1 – 2(.1841) = .6318| Thus, the probability is only .6318 that the sample mean will be within .3 ounce of the true population mean.
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