# Yacht Construction Essays Examples

Type of paper: Essay

Topic: Project, Probability, Standard, Completion, Percentage, Time, Organization, Deviation

Pages: 1

Words: 275

Published: 2020/10/08

## Business

Concept

The concept behind estimating project times is the understanding that the possible durations for a project to be completed form a normal distribution. The variance of the project is the square of the standard deviation of the time taken for the project. Once the standard deviation is obtained, the Z Score (the standard normal variate) can be obtained, from where the probability of completion of project in a given time can be derived using the normsdist function in Excel (Countryman, 2013).

## Problem

For the given problem, Tcp = expected project time as per critical path = 21 months.

Project variance = σt2 = 4 months. So, project standard deviation = σt = 2 months.

## Probability of Project Completion in 17 months or less

T1 = 17 months.

Z = (T1 - Tcp )/ σt = (17-21)/2 = -4/2 = -2.

This means that in a standard normal distribution, the probability of the project being completed in 17 months is two standard deviations less than the mean.

In terms of percentage, the probability is arrived at by using the formula: normsdist(z) = normsdist(-2) = 0.022750132, or 0.0228. The fraction, converted to percentage, is 0.0228 x 100 %= 2.28%.

Thus, we can say that the probability of the work being completed within 17 months is 2.28%. Probability of Project Completion in 20 months or less

T1 = 20 months.

Z = (T1 - Tcp )/ σt = (20-21)/2 = -1/2 = -0.5, ie 0.5 standard deviations less than the mean.

Probability of the project being completed in 20 months or less = normsdist(z) = normsdist(-0.5) = 0.3085. The fraction, converted to percentage, is 0.3085 x 100 %= 30.85%

Thus, we can say that the probability of the work being completed within 20 months is 30.85%.

## Probability of Project Completion in 23 months or less

T1 = 23 months.

Z = (T1 - Tcp )/ σt = (23-21)/2 = 2/2 = 1, ie 1 standard deviation more than the mean.

Probability of the project being completed in 20 months or less = normsdist(z) = normsdist(1) = 0.8413. The fraction, converted to percentage, is 0.8413 x 100 %= 84.13%

Thus, we can say that the probability of the work being completed within 23 months is 84.13%.

## Probability of Project Completion in 25 months or less

T1 = 25 months.

Z = (T1 - Tcp )/ σt = (25-21)/2 = 4/2 = 2, ie 2 standard deviations more than the mean.

Probability of the project being completed in 25 months or less = normsdist(z) = normsdist(2) = 0.9772. The fraction, converted to percentage, is 0.9772 x 100 %= 97.72%

Thus, we can say that the probability of the work being completed within 25 months is 97.72%.

## Application to Quantitative Analysis within an Organization

Probabilistic estimates can used to estimate the degree of risk and uncertainty involved in completing a project for an organization. The management can use probabilistic estimates to arrive at realistic timelines to complete a project. Organizations may be given incentives based on their effectiveness in completing projects within time, wherein the quantum of incentives could be linked to probabilities of completion within stipulated times. Thus, probabilistic scheduling can help organizations to be realistic and link performance to standardized metrics.

## Reference

Countryman, C. (2013). “PERT variance and probability of project completion.” [YouTube]. Retrieved 23 Jan 2015, from https://www.youtube.com/watch?v=1GGCLyOdCw8

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