Question

Assume that the duration of human pregnancies can be described by a Normal model with mean 268 days and standard deviation 14 days.

​a) What percentage of pregnancies should last between 274 and 281 days? (Round to two decimal places as needed.)

​b) At least how many days should the longest 25​% of all pregnancies​ last? (Round to one decimal place as needed.)

​c) Suppose a certain obstetrician is currently providing prenatal care to 58 pregnant women. Let y overbar represent the mean length of their pregnancies. According to the Central Limit​ Theorem, what's the distribution of this sample​ mean, y overbar​? Specify the​ model, mean, and standard deviation.

A. A normal model with mean ___ and standard deviation ___. (Type integers or decimals rounded to two decimal places as needed.)

B. A binomial model with ___ trials and a probability of success of ___. (Type integers or decimals rounded to two decimal places as needed.)

C. There is no model that fits this distribution.

​d) What's the probability that the mean duration of these​ patients' pregnancies will be less than

258 days?

- The probability that the mean duration of these patients' pregnancies will be less than 258 days is ___. (Round to four decimal places as needed.)

​days?

Answer 1

Here' the answer to the question. Please let me know in case you've doubts.

The normal distribution parameters are given below:

Mean = 268 days
Stdev = 14 days

Formula to standardize distribution = Z = (X-Mean)/Stdev

a. P(274<X<281)
Standardizing,

P( (274-268)/14 <Z< (281-268)/14)
=P(0.4286 <Z< 0.9286)
= 0.8234 - 0.6659
= 0.16

b. We have been asked to calculate a value "c" such that :
P(X>=c) = .25

So,
P(X<c) = 1-P(X>=c) = 1-.25 = .75
Standardizing:
P(Z< (c-268)/14) = .75
Z is 0.6745 for a cumulative prob of .75, Hence:
(c-268)/14 = 0.6745
c = 14*0.6745 + 268 = 277.443

Hence, answer to this part is 277.443

C. According to Central Limit theorm the correct answer is :

A. A normal model with mean = Population men (268) and Standard deviation = Population devition / sqrt(sample size) = 14/sqrt(58) = 1.84
We have used a property of CLT which approximates the sampling distribution' parameters using the population distribution and the sample size.

Answer: A. A normal model with mean 268 and standard deviation 1.84.

d.

P(X<258) = ?
Standardizing,
P(Z< (258-268)/14)
= P(Z< -0.7143)
= 0.2375

Answer is .2375