Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 12 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 38 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 = $\mu$ = 264

standard deviation = $\sigma$ = 12

Middle 50%

$\alpha$ = 1 - 50%  

$\alpha$ = 1 - 0.50 = 0.50

$\alpha$/2 = 0.25

1 - $\alpha$ /2 = 1 - 0.25 = 0.75

Z$\alpha$/2 = Z0.25 = -0.674

Z 1 $\alpha$ /2 = Z 0.75 = 0.674

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.674 * 12 +264

x = 255.912

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 0.674 * 12 + 264

x = 272.088

Between 256.9 and 272.1

( b )

n = 38

$\mu$ = 264

T

$\sigma$
T
= $\sigma$ / $\sqrt$ n = 12 / $\sqrt$ 38 = 1.9467

Middle 50%

$\alpha$ = 1 - 50%  

$\alpha$ = 1 - 0.50 = 0.50

$\alpha$/2 = 0.25

1 - $\alpha$ /2 = 1 - 0.25 = 0.75

Z$\alpha$/2 = Z0.25 = -0.674

Z 1 $\alpha$ /2 = Z 0.75 = 0.674

Using z-score formula  

T
= z * $\sigma$
T
+ $\mu$
T
  

= -0.674 * 1.9467 + 264

T

= 262.69

T

Using z-score formula  

T
= z * $\sigma$
T
+ $\mu$
T
  

= 0.674 * 1.9467 + 264

T

= 265.31

T

Between 262.7 and 265.3