Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 14 days.

In what range would you expect to find the middle 50% of most pregnancies?
Between  and .

If you were to draw samples of size 60 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample?
Between  and .

Enter your answers as numbers. Your answers should be accurate to 1 decimal places.

Answer 1

Solution:-

Given that,

mean = = 264 days

standard deviation = = 14 days

a) Using standard normal table,

P( -z < Z < z) = 50%

= P(Z < z) - P(Z <-z ) = 0.50

= 2P(Z < z) - 1 = 0.50

= 2P(Z < z) = 1 + 0.50

= P(Z < z) = 1.50 / 2

= P(Z < z) = 0.75

= P(Z < 0.6745) = 0.75

= z  ± 0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 14 + 264  

x = 254.6

Using z-score formula,

x = z * +

x = 0.6745 * 14 + 264

x = 273.4

The middle 50% are from 254.6 to 273.4 days

b) n = 60

= = 264 days

= / n = 14 / 60 = 1.81

Using z-score formula  

= z * +

= -0.6745 * 1.81 + 264

= 262.8

Using z-score formula  

= z * +

= 0.6745 * 1.81 + 264

= 265.2

The middle 50% are from 262.8 to 265.2 days