Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 263 days and a standard deviation of 13 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 50 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 .

Answer 1

Given that,

mean = = 263

standard deviation = = 13

middle 50% of score is

P(-z < Z < z) = 0.50

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

2 P(Z < z) - 1 = 0.50

2 P(Z < z) = 1 + 0.50 = 1.50

P(Z < z) = 1.50 / 2 = 0.75

P(Z < 0.67) = 0.75

z  ± 0.67 using z table

Using z-score formula  

x= z * +

x= 0.67*13+263

x= 271.71

x= z * +

x= -0.67*13+263

x= 254.29

Between 271.71 and 254.29

if we rounded.

Between 272 and 254 days

If you were to draw samples of size 50 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 .

solution:

Given that,

mean = = 263

standard deviation = = 13

n=50

= 263

=  / n= 13/ 50=1.84

middle 50% of score is

P(-z < Z < z) = 0.50

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

2 P(Z < z) - 1 = 0.50

2 P(Z < z) = 1 + 0.50 = 1.50

P(Z < z) = 1.50 / 2 = 0.75

P(Z < 0.67) = 0.75

z  ± 0.67 using z table

Using z-score formula  

x= z * +

x= 0.67*1.84+263

x= 264.23

x= z * +

x= -0.67*1.84+263

x= 261.77