To find the expected number of samples with a mean of 258 days or more from 100 independent random samples of size \( n = 14 \), we first calculate the probability \( P(X \geq 258) \).
Using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{258 - 266}{16 / \sqrt{14}} \approx -1.8708
\]
The probability is given by:
\[
P(X \geq 258) = 1 - \Phi(Z) = 1 - \Phi(-1.8708) \approx 0.9693
\]
Thus, the expected number of samples with a mean of 258 days or more is:
\[
E = 100 \times P(X \geq 258) \approx 100 \times 0.9693 \approx 3.07
\]
Next, we calculate the expected number of samples with a mean of 258 days or less. This is simply:
\[
P(X \leq 258) = \Phi(Z) \approx \Phi(-1.8708) \approx 0.0307
\]
Thus, the expected number of samples with a mean of 258 days or less is:
\[
E = 100 \times P(X \leq 258) \approx 100 \times 0.0307 \approx 96.93
\]
For a random sample of size \( n = 29 \), we calculate the probability that the mean gestation period is 258 days or less.
Using the Z-score formula:
\[
Z = \frac{258 - 266}{16 / \sqrt{29}} \approx -2.6926
\]
The probability is given by:
\[
P(X \leq 258) = \Phi(Z) \approx \Phi(-2.6926) \approx 0.0035
\]
If we were to take 100 independent random samples of size \( n = 29 \), we would expect:
\[
E = 100 \times P(X \leq 258) \approx 100 \times 0.0035 \approx 0.35
\]
- Part B: Expected number of samples with mean 258 days or more: \( \boxed{3} \)
- Part C: Expected number of samples with mean 258 days or less: \( \boxed{97} \)
- Part D: Probability that a sample of 29 pregnancies has a mean of 258 days or less: \( \boxed{0.0035} \)