In general, length of pregnancies are normally distributed with a mean of 268 da
ID: 3129534 • Letter: I
Question
In general, length of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.
a. One classic use of the normal distribution is inspired by a letter to “Dear Abby” in which a wife claimed to have given birth 306 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting 306 days or longer. What does the result suggest?
Although it is not required to use the rare event rule to answer the above question, provide correct assumption, correct observe event, correct reasoning for calling the observe event as rare or not so rare under your assumption, and finally, a correct conclusion. Write the answer in a paragraph form.
b. A baby is considered premature if the length of pregnancy is in the lowest 8%. Find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 306
u = mean = 268
s = standard deviation = 15
Thus,
z = (x - u) / s = 2.533333333
Thus, using a table/technology, the right tailed area of this is
P(z > 2.533333333 ) = 0.005649173
Here, the length of all pregnancies is assumed to be normally distirbuted, and we "observed" an event of 306 days of pregnancy. It has very small probability, 0.0056, which is less than 0.05. Hence, as this is very rare, then this report of pregnancy may actually be false.
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b)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.08
Then, using table or technology,
z = -1.40507156
As x = u + z * s,
where
u = mean = 268
z = the critical z score = -1.40507156
s = standard deviation = 15
Then
x = critical value = 246.9239266 days [ANSWER]