Question

Assume that the duration of human pregnancies can be described by a Normal model with mean 267 days and standard deviation 14 days.

​a) What percentage of pregnancies should last between 272 and 284 days?

​b) At least how many days should the longest 20​%of all pregnancies​ last?

​c) Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let y

represent the mean length of their pregnancies. According to the Central Limit​ Theorem, what's the distribution of this sample​ mean, y​? Specify the​ model, mean, and standard deviation.

​d) What's the probability that the mean duration of these​ patients' pregnancies will be less than 256 ​days?

Answer 1

Part a)

X ~ N ( µ = 267 , σ = 14 )
P ( 272 < X < 284 )

P (( 272 - 267 ) / 14 < ( X - µ ) / σ < ( 284- 267 ) / 14 )
= P ( Z < 1.07) - P ( Z < 0.36 )

value of z is obtain from stnadard normal table
= 0.8580 - 0.6395
P ( 272 < X < 284 ) = 0.2185

# 21.85%  of pregnancies should last between 272 and 284 days

Part b)

X ~ N ( µ = 267 , σ = 14 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.2 = 0.8
To find the value of x
Looking for the probability 0.8 in standard normal table to calculate Z score = 0.8416
Z = ( X - µ ) / σ
0.8416= ( X - 267 ) / 14
X = 278.78
P ( X > 278.78 ) = 0.8

# At least 279 days should the longest 20​%of all pregnancies​ last

Part c)

µX̅ = µ = 267
σX̅ = σ / √ (n) = 14/√60 = 1.80739

Shape would be Normal with mean = 267 and standard deviation = 1.80739

Part d)

X ~ N ( µ = 267 , σ = 14 )
P ( X < 256 )
P(( X - µ ) / (σ/√(n)< ( 256 - 267 ) / ( 14 / √60 ))
P ( X < 256 ) = P ( Z < -6.09 )
P ( X̅ < 256 ) = 0