Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 270 days and a standard deviation of 15 days.

In what range would you expect to find the middle 98% of most pregnancies?
Between _____and ____.

If you were to draw samples of size 36 from this population, in what range would you expect to find the middle 98% 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.

Please only respond to this question if you can do so accurately and entirely.

Answer 1

Solution :

mean = = 270

standard deviation = = 15

Using standard normal table,

(1)

P(Z > z) = 98%

1 - P(Z < z) = 0.98

P(Z < z) = 1 - 0.98 = 0.02

P(Z < -2.054) = 0.02

z = - 2.054

x = z * +

x = -2.054 * 15 + 270 = 239.2 and

P(Z < z) = 0.98

P(Z < 2.054) = 0.98

z = 2.054

x = z * +

x = 2.054 * 15 + 270 = 300.8

Between 239.2 and 300.8

(2)

n = 36

= 270

= 15 / 36 = 15 / 6 = 2.5

P(Z > z) = 98%

1 - P(Z < z) = 0.98

P(Z < z) = 1 - 0.98 = 0.02

P(Z < -2.054) = 0.02

z = - 2.054

= z * + = -2.054 * 2.5 + 270 = 264.9 and

P(Z < z) = 0.98

P(Z < 2.054) = 0.98

z = 2.054

= z * + = 2.054 * 2.5 + 270 = 275.1

Between 264.9 and 275.1 .