Given:
- Project duration (mean, μ\mu μ) = 20 weeks
- Standard deviation (σ\sigma σ) = 4 weeks
- We want to find xxx such that there is a certain chance the project will be completed in xxx weeks or less.
This is a problem of finding a value xxx on a normal distribution curve with mean 20 and standard deviation 4. The question is incomplete about the exact probability or confidence level, but typically, if we want to find xxx corresponding to a certain probability, we use the Z-score formula:
Z=x−μσZ=\frac{x-\mu}{\sigma}Z=σx−μ
If the question implies the chance is 50%, then x=μ=20x=\mu =20x=μ=20 weeks. If the question implies the chance is about 84% (one standard deviation above the mean), then:
x=μ+σ=20+4=24weeksx=\mu +\sigma =20+4=24\quad \text{weeks}x=μ+σ=20+4=24weeks
If the question implies the chance is about 16% (one standard deviation below the mean), then:
x=μ−σ=20−4=16weeksx=\mu -\sigma =20-4=16\quad \text{weeks}x=μ−σ=20−4=16weeks
Since the answer choices are 18.4, 0.8, and 21, the closest reasonable xxx value for a probability less than 50% but close to it would be around 18.4 weeks. To check this, calculate the Z-score for x=18.4x=18.4x=18.4:
Z=18.4−204=−1.64=−0.4Z=\frac{18.4-20}{4}=\frac{-1.6}{4}=-0.4Z=418.4−20=4−1.6=−0.4
From the standard normal distribution table, a Z of -0.4 corresponds to about 34% chance to be less than 18.4 weeks. Similarly, x=21x=21x=21 weeks gives:
Z=21−204=0.25Z=\frac{21-20}{4}=0.25Z=421−20=0.25
Which corresponds to about 60% chance. Since the question is vague but asks for the closest answer for xxx, the best choice is 18.4 weeks , which is a plausible project completion time less than the mean, reflecting a chance less than 50% but close to it. Answer: 18.4 weeks
