Question

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Explanation:

Identify given parameters

Using the Normal Distribution knowledge point, we define the random variable \(X\) as the pregnancy length.

• Mean: \(\mu\)
• Standard deviation: \(\sigma = 11\) days
• Target value: \(X > \mu + 22\) days

Calculate the Z-score

Using the Normal Distribution knowledge point

$$ z = \frac{x - \mu}{\sigma} = \frac{(\mu + 22) - \mu}{11} = \frac{22}{11} = 2 $$

Apply the Empirical Rule

Using the Empirical Rule knowledge point

• Approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean (\(-2 < Z < 2\)).
• The remaining area in both tails is \(100\% - 95\% = 5\%\).
• Since the normal distribution is symmetric, the area in the upper tail (\(Z > 2\)) is:

$$ P(Z > 2) = \frac{5\%}{2} = 2.5\% $$

Answer:

Actual lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length with a standard deviation of 11 days. About what percentage to occur more than 22 days after the mean pregnancy length?

About <blank>2.5</blank>% of births would be expected to occur more than 22 days after the mean pregnancy length.
(Type an integer or a decimal.)