To solve this problem using the PERT (Program Evaluation Review Technique) formulas, let's find the expected time and the variance for activity \( C \).
### Step-by-Step Solution
a) Expected (estimated) time \( t_e \) for activity \( C \)
The PERT formula for expected time \( t_e \) is:
[tex]\[ t_e = \frac{a + 4m + b}{6} \][/tex]
Given:
- Optimistic time (\( a \)) = 9 weeks
- Most likely time (\( m \)) = 12 weeks
- Pessimistic time (\( b \)) = 18 weeks
Substitute the values into the formula:
[tex]\[ t_e = \frac{9 + 4(12) + 18}{6} \][/tex]
Calculate inside the parentheses first:
[tex]\[ t_e = \frac{9 + 48 + 18}{6} \][/tex]
[tex]\[ t_e = \frac{75}{6} \approx 12.5 \][/tex]
So, the expected (estimated) time for activity \( C \) is 12.5 weeks.
b) Variance for activity \( C \)
The PERT formula for the variance \( \sigma^2 \) is:
[tex]\[ \text{Variance} = \left( \frac{b - a}{6} \right)^2 \][/tex]
Given:
- Optimistic time (\( a \)) = 9 weeks
- Pessimistic time (\( b \)) = 18 weeks
Substitute the values into the formula:
[tex]\[ \text{Variance} = \left( \frac{18 - 9}{6} \right)^2 \][/tex]
Calculate inside the parentheses first:
[tex]\[ \text{Variance} = \left( \frac{9}{6} \right)^2 \][/tex]
[tex]\[ \text{Variance} = \left( 1.5 \right)^2 \][/tex]
[tex]\[ \text{Variance} = 2.25 \][/tex]
So, the variance for activity [tex]\( C \)[/tex] is 2.25 weeks.
