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Part 6 of 6

Ross Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in the following table for building new motels.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\multirow[b]{2}{}{Activity} & \multicolumn{3}{|c|}{Time (weeks)} & \multirow{2}{}{\begin{tabular}{l}
Immediate \\
Predecessor(s)
\end{tabular}} & \multirow[b]{2}{}{Activity} & \multicolumn{3}{|c|}{Time (weeks)} & \multirow{2}{}{\begin{tabular}{l}
Immediate \\
Predecessor(s)
\end{tabular}} \\
\hline
& [tex]$a$[/tex] & [tex]$m$[/tex] & [tex]$b$[/tex] & & & [tex]$a$[/tex] & [tex]$m$[/tex] & [tex]$b$[/tex] & \\
\hline
A & 4 & 9 & 10 & - & G & 3 & 3 & 4 & [tex]$\overline{C, E}$[/tex] \\
\hline
B & 2 & 9 & 24 & A & H & 2 & 2 & 2 & F \\
\hline
C & 9 & 12 & 18 & A & I & 5 & 5 & 5 & F \\
\hline
D & 4 & 7 & 10 & A & J & 6 & 7 & 14 & [tex]$\overline{D, G, H}$[/tex] \\
\hline
E & 1 & 3 & 4 & B & K & 1 & 1 & 4 & [tex]$\overline{I, J}$[/tex] \\
\hline
F & 5 & 8 & 20 & [tex]$\overline{C, E}$[/tex] & & & & & \\
\hline
\end{tabular}

a) The expected (estimated) time for activity [tex]$C$[/tex] is 12.5 weeks. (Round your response to two decimal places.)

b) The variance for activity [tex]$C$[/tex] is 2.25 weeks. (Round your response to two decimal places.)

c) Based on the calculation of the estimated times, the critical path is [tex]$A-B-E-F-H-J-K$[/tex].

d) The estimated time for the critical path is 42.5 weeks. (Round your response to two decimal places.)

e) The activity variance along the critical path is 22.97 weeks. (Round your response to two decimal places.)

f) Using the standard normal table, the probability that Hopkins Hospitality will finish the project in 42 weeks or less is [tex]$\square$[/tex]. (Enter as a probability and round your response to three decimal places.)



Answer :

GinnyAnswer
To find the probability that Hopkins Hospitality will finish the project in 42 weeks or less, we need to follow a systematic approach using the given parameters and some statistical calculations outlined below.

First, we highlight the given data:
- Estimated time \(T_{CP}\) for the critical path is 42.5 weeks.
- Variance \(\sigma^2_{CP}\) along the critical path is 22.97 weeks.
- Target project completion time \(T\) is 42 weeks.

### Step-by-step Solution:

1. Calculate the Standard Deviation (\(\sigma_{CP}\)) of the Project Time:
The standard deviation is the square root of the variance. This helps in understanding the spread of the possible completion times around the mean.

[tex]\[ \sigma_{CP} = \sqrt{\sigma^2_{CP}} = \sqrt{22.97} \approx 4.793 \text{ weeks} \quad (\text{rounded to three decimal places}) \][/tex]

2. Calculate the Z-score:
The Z-score is a measure that describes how many standard deviations a particular value is from the mean. It is calculated as:

[tex]\[ Z = \frac{T - T_{CP}}{\sigma_{CP}} \][/tex]

Substituting the given values, we get:

[tex]\[ Z = \frac{42 - 42.5}{4.793} \approx -0.104 \quad (\text{rounded to three decimal places}) \][/tex]

3. Determine the Probability:
The probability that the project will complete in 42 weeks or less can be found using the Cumulative Distribution Function (CDF) for the normal distribution at the computed Z-score.

Using a standard normal distribution table or a computational tool, we find the cumulative probability corresponding to the Z-score of -0.104.

[tex]\[ P(Z < -0.104) \approx 0.458 \quad (\text{rounded to three decimal places}) \][/tex]

### Conclusion:
The probability that Hopkins Hospitality will finish the project in 42 weeks or less is approximately:

[tex]\[ \boxed{0.458} \][/tex]

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