Answer :
Answer:
0.9599 is the probability that the project will be completed before the late-payment deadline.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 145
Variance, [tex]\sigma^2[/tex] = 400
Standard deviation:
[tex]\sigma = \sqrt{400} = 20[/tex]
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P( project will be completed before the late-payment deadline)
We have to evaluate
[tex]P( x \leq 180) = P( z \leq \displaystyle\frac{180 - 145}{20}) = P(z \leq 1.75)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \leq 180) =0.9599= 95.99\%[/tex]
0.9599 is the probability that the project will be completed before the late-payment deadline.
The probability that the project will be completed before the late-payment deadline is 0.9599
Given that,
- Estimated project time of 145 working days with a project variance of 400.
Calculation:
Since the variance is 400 so the standard deviation is 20.
Now
The probability is
[tex]= P (z \leq \frac{180-45}{20} ) = P(z\leq 1.75)[/tex]
So,
[tex]P(z\leq 180) = 0.9599 = 95.99\%[/tex]
Learn more about the probability here: https://brainly.com/question/795909?referrer=searchResults