Final answer:
To find the duration of pregnancy for the shortest 20%, we use the 20th percentile z-score of -0.84. Applying the z-score formula, we discover that a pregnancy lasting 264.44 days falls into this percentile. This calculation is based on a normal distribution with a mean of 272 days and a standard deviation of 9 days.
Step-by-step explanation:
To determine how many days a pregnancy would last for the shortest 20% given a mean of 272 days and a standard deviation of 9 days, we can use the concept of the normal distribution and z-scores. The shortest 20% corresponds to the 20th percentile in a normal distribution.
First, we find the z-score that corresponds to the 20th percentile using a standard normal distribution table or calculator. For the 20th percentile, the z-score is approximately -0.84.
Next, we use the formula for converting a z-score into the original score (X):
X = μ + (z × σ)
Where μ is the mean, σ is the standard deviation, and z is the z-score.
For this problem:
- Mean (μ) = 272 days
- Standard deviation (σ) = 9 days
- Z-score (z) = -0.84
Using the formula:
X = 272 + (-0.84 × 9)
X = 272 - 7.56
X = 264.44 days
Therefore, a pregnancy would last 264.44 days for the shortest 20%, which is rounded to 264.44 days when rounded to two decimal places.