a) To find the probability of a pregnancy lasting 308 days or longer, we can use the properties of the normal distribution.
Given that the length of pregnancies is normally distributed with a mean of 268 days and a standard deviation of 15 days, we can calculate the z-score for 308 days using the formula:
z = (x - μ) / σ
where:
- x is the value we are interested in (308 days),
- μ is the mean (268 days), and
- σ is the standard deviation (15 days).
Plugging in the values, we get:
z = (308 - 268) / 15 = 2.6667
Next, we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. From the table, we find that the probability is approximately 0.9960.
This means that the probability of a pregnancy lasting 308 days or longer is approximately 0.9960, or 99.60%.
The result suggests that a pregnancy lasting 308 days or longer is quite rare, occurring only about 0.40% of the time. It may be considered an outlier or an unusual occurrence.
b) To find the length that separates premature babies from those who are not premature, we need to determine the value corresponding to the lowest 4% of the distribution.
Since the distribution is symmetric, we can find the z-score that corresponds to the lower 4% and then convert it back to the length using the formula:
z = (x - μ) / σ
To find the z-score, we look up the area of 0.04 in the standard normal distribution table, which corresponds to the lower 4%. The z-score is approximately -1.750.
Now we can solve for x using the formula:
-1.750 = (x - 268) / 15
Solving for x, we get:
-1.750 * 15 + 268 = x
x ≈ 241.25
Therefore, the length that separates premature babies from those who are not premature is approximately 241.25 days.
This result can be helpful to hospital administrators in planning for the care of premature babies. It allows them to identify and allocate resources for the special care that premature babies often require.