Question

The lengths of pregnancies in a small rural village are Normally distributed with a mean of 264 days and a standard deviation of 15 days.

We expect that 99.7% of pregnancies will last
between  and  days.

Answer 1

X ~ N ( µ = 264 , σ = 15 )
P ( a < X < b ) = 0.997
Dividing the area 0.997 in two parts we get 0.997/2 = 0.4985
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.4985
Area above the mean is b = 0.5 + 0.4985
Looking for the probability 0.0015 in standard normal table to calculate Z score = -2.9677
Looking for the probability 0.9985 in standard normal table to calculate Z score = 2.9677
Z = ( X - µ ) / σ
-2.9677 = ( X - 264 ) / 15
a = 219.4845
2.9677 = ( X - 264 ) / 15
b = 308.5155
P ( 219.4845 < X < 308.5155 ) = 0.997