Questions: The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy last at least 300 days? Round to four decimal places. A. 0.0164 B. 0.0179 C. 0.9834 D. 0.4834

Solution Steps

We need to find the probability that a pregnancy lasts at least 300 days, given that the lengths of human pregnancies are normally distributed with a mean (\( \mu \)) of 268 days and a standard deviation (\( \sigma \)) of 15 days.

To find the probability, we first calculate the Z-score for \( X = 300 \) days using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ Z = \frac{300 - 268}{15} = \frac{32}{15} \approx 2.1333 \]

The probability that a pregnancy lasts at least 300 days can be expressed as:

\[ P(X \geq 300) = 1 - P(X < 300) = 1 - \Phi(Z) \]

Where \( \Phi(Z) \) is the cumulative distribution function (CDF) of the standard normal distribution. Thus, we have:

\[ P(X \geq 300) = 1 - \Phi(2.1333) \]

Using the standard normal distribution table or calculator, we find:

\[ \Phi(2.1333) \approx 0.9836 \]

Therefore:

\[ P(X \geq 300) = 1 - 0.9836 = 0.0164 \]

Final Answer