Assume that the distribution of the duration of human pregna

Assume that the distribution of the duration of human pregnancies can be approximated with a normal distribution with a mean of 266 days and a standard deviation of 16 days. What percentage of pregnancies should last between 270 and 280 days? Find a value x such that 25% of the pregnancies of a duration that is longer than x days. We select 10 pregnant women at random. What is the probability that the average duration of these 10 pregnancies will be less than 260 days?

Solution

Normal Distribution
Mean ( u ) =266
Standard Deviation ( sd )=16
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 270) = (270-266)/16
= 4/16 = 0.25
= P ( Z <0.25) From Standard Normal Table
= 0.59871
P(X < 280) = (280-266)/16
= 14/16 = 0.875
= P ( Z <0.875) From Standard Normal Table
= 0.80921
P(270 < X < 280) = 0.80921-0.59871 = 0.2105 ~ 21.05%                  

b)
P ( Z > x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is 0.67
P( x-u/ (s.d) > x - 266/16) = 0.25
That is, ( x - 266/16) = 0.67
--> x = 0.67 * 16+266 = 276.784                  

c)
P(X < 260) = (260-266)/16/ Sqrt ( 10 )
= -6/5.0596= -1.1859
= P ( Z <-1.1859) From Standard NOrmal Table
= 0.1178   ~ 11.78%