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.

In what range would we expect to find the middle 68% of most lengths of pregnancies? Round the answer to one decimal place.

Between and days

If we were to draw samples of size 56 from this population, in what range would we expect to find the middle 68% of most averages for the lengths of pregnancies in the sample? Round the answer to one decimal place.

Between and days

Answer 1

Solution :

Given that,

mean = = 264

standard deviation = = 15

Using standard normal table,

a )P(-z < Z < z) = 68%
P(Z < z) - P(Z < z) = 0.68
2P(Z < z) - 1 = 0.68
2P(Z < z ) = 1 + 0.68
2P(Z < z) = 1.68
P(Z < z) = 1.68 / 2
P(Z < z) = 0.75
-z =- 0.99 and z =0.99

Using z-score formula,

x = z * +

x = - 0.99 * 15+ 264

x = 249 days

x = z * +

x = 0.99 * 15+ 264

x = 279 days

b ) n =56

= 264

= / n =15 56= 2.004

P(-z < Z < z) = 68%
P(Z < z) - P(Z < z) = 0.68
2P(Z < z) - 1 = 0.68
2P(Z < z ) = 1 + 0.68
2P(Z < z) = 1.68
P(Z < z) = 1.68 / 2
P(Z < z) = 0.75
-z =- 0.99 and z =0.99

Using z-score formula,

= z * +

= - 0.99 * 2.004+ 264

= 262 days

= z * +

= 0.99 * 2.004+ 264

=266 days