Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation of 13 days.

In what range would we expect to find the middle 95% of most lengths of pregnancies? Then Round the answer to one decimal place to find the answer..

Between ? and ? days.

If we were to draw samples of size 38 from this population, in what range would we expect to find the middle 95% of most averages for the lengths of pregnancies in the sample? Then Round the answer to one decimal place to find the answer. .

Between ? and ? days.

Answer 1

Given that,

mean = = 266

standard deviation = = 13

middle 95% of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0. 95= 1.95

P(Z < z) = 1.95/ 2 = 0.975

P(Z <1.96 ) = 0.975

z  ±1.96   

Using z-score formula  

x= z * +

x=-1.96 *13+266

x= 240.5

z = 1.96

Using z-score formula  

x= z * +

x= 1.96 *13+266

x= 291.5

(b)

given that,

mean = = 266

standard deviation = = 13

n=38

= 266

= / n = 13/ 38= 2.1089

middle 95% of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0. 95= 1.95

P(Z < z) = 1.95/ 2 = 0.975

P(Z <1.96 ) = 0.975

z  ±1.96   

Using z-score formula  

x= z * +

x=-1.96 *2.1089+266

x= 261.9

z = 1.96

Using z-score formula  

x= z * +

x= 1.96 *2.1089+266

x= 270.1