Assume that the duration of human pregnancies can be described by a Normal model
ID: 3128222 • Letter: A
Question
Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and standard deviation 16 days. What percent of pregnancies should last between 270 and 280 days? At least how many days should the longest 25% of all pregnancies last? Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let ¯X represent the mean length of their pregnancies. According to the Central Limit Theorem, what is the distribution of the sample mean, ¯X ? Specify the model, mean and standard deviation. Justify your answer (i.e. check conditions!) What is the probability that the mean duration of these patients’ pregnancies will be less than 260 days? The duration of human pregnancies may not actually follow a Normal model, as described. Explain why the actual distribution may be somewhat skewed to the left. If the correct model is in fact skewed, does that change your answers to parts a), b) and c)? Explain why or why not for each.
Explanation / Answer
Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and standard deviation 16 days. What percent of pregnancies should last between 270 and 280 days?
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 270
x2 = upper bound = 280
u = mean = 266
s = standard deviation = 16
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.25
z2 = upper z score = (x2 - u) / s = 0.875
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.598706326
P(z < z2) = 0.809213047
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.210506721 = 21.05% [ANSWER]
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At least how many days should the longest 25% of all pregnancies last?
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.25 = 0.75
Then, using table or technology,
z = 0.67448975
As x = u + z * s,
where
u = mean = 266
z = the critical z score = 0.67448975
s = standard deviation = 16
Then
x = critical value = 276.791836 [ANSWER]
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According to the Central Limit Theorem, what is the distribution of the sample mean, ¯X ?
Specify the model, mean and standard deviation.
It is normally distributed, with the same mean, 266 days.
It has a reduced standard deviation, 16/sqrt(60) = 2.065591118.
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What is the probability that the mean duration of these patients’ pregnancies will be less than 260 days?
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 260
u = mean = 266
n = sample size = 60
s = standard deviation = 16
Thus,
z = (x - u) * sqrt(n) / s = -2.90473751
Thus, using a table/technology, the left tailed area of this is
P(z < -2.90473751 ) = 0.001837806 [ANSWER]
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The duration of human pregnancies may not actually follow a Normal model, as described. Explain why the actual distribution may be somewhat skewed to the left. If the correct model is in fact skewed, does that change your answers to parts a), b) and c)? Explain why or why not for each.
As we can see, they are pregnancy lengths, and of course, very few pregnancies become very short, and usually it is around 9 months. Hence, it probably is skewed to the left.
For parts A and B, the answer would be affected as we assume that they have a normal distribution.
However, part c would not be changed, because by central limit theorem, sampling distribution of means will be normally distributed for large enough sample size.
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