• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Simulation

Extracts from this document...

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)
...read more.

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.
...read more.

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.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Tide Modeling

    Finally Graphing Package was used in order to graph the data and so that it could be compare to the sine model Once that was done in order to better compare the model with the best- fit graph it was decided to plug in values for ??

  2. Mathematics (EE): Alhazen's Problem

    is a solution between the points 11 and 12 and also between the points 6 and 7 on the circumference, by looking at the graph one can see that these chords leading to the points are in fact the same chords and the points would therefore definitely not work as

  1. A logistic model

    A harvest of 8x103 fish is made per annum. Year (after Population Year Population attaining stability) 1 3.00?104 16 3.94?104 2 3.20?104 17 3.95?104 3 3.30?104 18 3.96?104 4 3.39?104 19 3.97?104 5 3.47?104 20 3.97?104 6 3.55?104 21 3.98?104 7 3.62?104 22 3.98?104 8 3.68?104 23 3.99?104 9 3.73?104

  2. This essay will examine theoretical and experimental probability in relation to the Korean card ...

    I will start by giving brief idea on what is probability and how it relates to the real life world situations. I will then start examining the chances of getting each hand with "Sut-Da" then compare with the experimental value.

  1. Networks - Konigsberg Bridge Problem.

    Not Traversable Not Traversable Above is the table with the nodes and if the network is traversable. From the table, I found out that when a network if traversable, then the network has two odd nodes or less (it is not possible to has one odd node).

  2. Creating a logistic model

    un If we let 10000 fish with an initial growth rate of 2.3 cultivate for 20 years, the population of growth would be: Year Population of Fish Growth Rate t un+1 = (-0.000026 � un + 2.56) un rn = -0.000026 � un + 2.56 0 10000 2.3 1 23000

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work