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The duration (in days) of human pregnancies following approximately Normal with μ = 266 and σ = 216. How many days would a human pregnancy need to last to be among the top 15% of all pregnancy duration.

User Noev
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Final answer:

To be in the top 15% of all human pregnancy durations, the pregnancy would need to last approximately 283 days. This calculation is based on the z-score for the top 85% of a standard normal distribution and the given mean and standard deviation of pregnancy durations.

Step-by-step explanation:

The question asks to determine the number of days a human pregnancy must last to be in the top 15% of all pregnancy durations. Given that the duration of human pregnancies is normally distributed with a mean (μ) of 266 days and a standard deviation (σ) of 16 days, we first need to find the z-score that corresponds to the top 15% of a standard normal distribution. Then, we can use the z-score to calculate the corresponding duration in days.

To find the z-score that cuts off the top 15% of the standard normal distribution, we can use a z-table or calculator. We look for the value that corresponds to the cumulative area of 0.8500 (since the top 15% is equivalent to 100% - 15% = 85%). This value is approximately z = 1.036.

With the z-score and the information on the mean and standard deviation of the pregnancy duration, we apply the formula:

Duration = μ + z × σ

Substituting the given values, we get:

Duration = 266 + 1.036 × 16

Duration ≈ 266 + 16.576

Duration ≈ 282.576 days

Therefore, to be in the top 15% of all pregnancy durations, the pregnancy would need to be approximately 283 days long.

User Thierry Lam
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