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Simulation

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Introduction

The manager of a wine bottling plant has received a special order for two wine bottles. The two wine bottles require corks with diameters between 2.7 centimeters and 3.3 centimeters to properly fit the opening of the bottle. From past experience, the manger knows that corks used at his bottling plant have diameters that are normally distributed with a mean of 3 centimeters and a standard deviation of 1 centimeter.

  1. Determine the probability of finding a cork of the appropriate size in plant cork bin selected at random. Round your answer to the nearest thousandth.
  1. A Normal Probability Distribution is symmetric, single-peaked, and bell shaped. The points of inflection lie between plus and minus one standard deviations from the mean. The total area under the curve is zero, and the empirical rule values should fit relatively well.
  2. To solve this problem, we are looking for the P (2.7 cm < x < 3.3 cm). To find this probability, we need to standardize the 2.7 cm and 3.3 cm. Do to so, we have to use the formula z = (xi – x)
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Middle

236 – 999 to stand for “no”—meaning that these corks are not between 2.7 centimeters and 3.3 centimeters.We are going to use a random digit table.
  1. Trial:
  1. A trial must be at least two tries so that it ensures that at least two corks are selected of the right size.
  2. For the trial to end, there must be two “yes”’s, so that there is two corks with the diameters desired.
  1. Observation of Interest:
  1. The diameters of the corks. Specifically, the corks with the diameters between 2.7 centimeters and 3.3 centimeters.
  1. Perform your simulation three times. (That is, run 3 trials of your simulation.) Start at the left most digit in the first row of the table and move across. Describe your simulation run in sufficient detail to make clear the reader. Demonstrate by marking your table.
  1. Refer to previous question for trial in detail.
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Conclusion

i – x) / s which is your data value minus the mean, and all of that divided by the standard deviation.
  1. Z-score for 1.8 = (1.8 – 3) / 1 = -1.2
  2. Z-score for 2.5 = (2.5 -3) / 1 = -0.5
  1. Once the z-score is determined, we use the z-score probability chart to determine the probability associated with the z-score.
  1. P (z = -1.2) = .1151
  2. P (z = -0.5) = .3085
  1. To find the probability with the associated interval, we subtract the probability from -1.2 from -0.5.
  1. .3085 - .1151 = .1934
  1. Rounded to the nearest thousandth place, P (1.8 cm < x < 2.5 cm) is approximately .193 or 19.3%.
  1. Compute an estimate of the expected number of corks that must be examined in the cork bin to find two corks between 2.7 cm and 3.3 cm in diameter. Use the appropriate distribution above.
  1. The best estimate of the expected number of corks is the sample mean of approximately 6.7, for distribution B. This can be arrived by using the formula:
  1. x = (∑ x * f) / n where f stands for frequency (or number of trials for this situation), n is the number of trial simulations performed in total, and x is the value on the x-axis for that particular bar.
  • (3*6) + (4*10) + (5*16) + (6*22) = (7*24) + (8*13) + (9*12) + (10*5) + (11*2) + (12*4) = 770.
  • 770 / 15 = 6.69565217391 which rounds to 6.7.

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