Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 13 days.

In what range would you expect to find the middle 68% of most pregnancies?
Between  and .

If you were to draw samples of size 31 from this population, in what range would you expect to find the middle 68% 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

Given that,

mean = = 261

standard deviation = = 13

middle 68% of score is

P(-z < Z < z) = 0.68

P(Z < z) - P(Z < -z) = 0.68

2 P(Z < z) - 1 = 0.68

2 P(Z < z) = 1 + 0.68 = 1.68

P(Z < z) = 1.68 / 2 = 0.84

P(Z < 0.99) = 0.84

z  ± 0.99 using z table

Using z-score formula  

x= z * +

x= 0.99*13+261

x= 273.87

x= z * +

x= -0.99*13+261

x= 248.13.

Between 274 and 248 days

If you were to draw samples of size 31 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample?
Between  and

solution:

Given that,

mean = = 261

standard deviation = = 13

n=31

= 261

=  / n= 13/ 31=2.33

middle 68% of score is

P(-z < Z < z) = 0.68

P(Z < z) - P(Z < -z) = 0.68

2 P(Z < z) - 1 = 0.68

2 P(Z < z) = 1 + 0.68 = 1.68

P(Z < z) = 1.68 / 2 = 0.84

P(Z < 0.99) = 0.84

z  ± 0.99 using z table

Using z-score formula  

Using z-score formula  

x= z * +

x= 0.99*2.33+261

x= 263

x= z * +

x= -0.99*2.33+261

x= 259

Between 263 and 259