The duration (in days) of human pregnancies follows approximately the Normal with mean 266 days and standard deviation 16 days. How many days would a human pregnancy need to last to be among the top 10% of all durations?

286.5 days.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 266 days

Standard deviation = σ = 16 days.

Let the number that corresponds to the top 10% be a.

P(x > a) = 0.1

Let the z-score of a be z'

z' = (x - μ)/σ = (a - 266)/16

Using the normal distribution table, we can obtain the z-score that corresponds to z'

P(x > a) = P(z > z') = 1 - P(z ≤ z') = 0.1

P(z ≤ z') = 1 - 0.1 = 0.9

From the tables, z' = 1.282

P(z ≤ 1.282) = 0.9

And P(z > 1.282) = 0.1

1.282 = (a - 266)/16

16 × 1.282 = a - 266

20.512 = a - 266

a = 286.512 days

Hope this Helps!!!

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step-by-step explanation: