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PIT_PIT [208]
3 years ago
13

mauricio estimates that the first module of the project could be completed in as few as 15 days or could take as many as 25 days

but most likely will requrie 20 days. what is the 20 day estimate called?
Mathematics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

<em>Most likely time, </em>according to PERT (Program evaluation and review technique).

Step-by-step explanation:

PERT is "a statistical tool used in <em>project management" (Program evaluation and review technique (2020), </em>in Wikipedia), and it is commonly used with CPM <em>(Critical Path Method)</em> to manage projects.

Inside PERT, there are different defined times to accomplished an activity in a project, that is:

  • An <em>optimistic time</em> or minimum time required to accomplished an activity, i.e., if everything goes better than normal, the activity is accomplished before expected.
  • A <em>pessimistic time, </em>a time quite the opposite to optimistic time.
  • A <em>most likely time</em>, or a time required to accomplished an activity if everything goes as expected or normally.
  • An <em>expected time</em>, an statistical estimation.

Considering the question, we have that the <em>time</em> when "the first module of the project could be completed":

  • "[...] in as few as 15 days"  is the <em>optimistic time</em>.
  • "[...] or could take as many as 25 days" is the <em>pessimistic time</em>.
  • "[...] but most likely will require 20 days" is the <em>most likely time</em>.

As a result, the <em>20-day estimate</em> is called the <em>most likely time</em> in the context of the PERT/CPM techniques.

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4 0
3 years ago
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5) Two machines M1, M2 are used to manufacture resistors with a design
Basile [38]

Answer:

Since M1 has the higher probability of being in the desired range, we choose M1.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.

So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.

For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.

M1:

Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that \mu = 1050, \sigma = 100

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

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

Z = \frac{1100 - 1050}{100}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 900

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

Z = \frac{900 - 1050}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.6915 - 0.0668 = 0.6247.

M1 has a 62.47% probability of being in the desired range.

M2:

M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that \mu = 1000, \sigma = 120

X = 1100

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

Z = \frac{1100 - 1000}{120}

Z = 0.83

Z = 0.83 has a pvalue of 0.7967.

X = 900

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

Z = \frac{900 - 1000}{120}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

0.7967 - 0.2033 = 0.5934

M2 has a 59.34% probability of being in the desired range.

Since M1 has the higher probability of being in the desired range, we choose M1.

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Answer:

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Step-by-step explanation:

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