Question

6.

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Use this information to compute the answers for the following questions:

a.Find the probability that a randomly selected pregnancy is less than250 days.Round to 3 decimals.

b.Find the probability that a randomly selected pregnancy is between250 and 280days.Round to 3 decimals.

c.Find the pregnancy duration(in days) that separates the bottom 90% of pregnancies fromthe top 10%.Roundto the nearestwholeday.

Answer 1

Solution :

Given that,

mean = = 268

standard deviation = = 15

a ) P( x < 250 )

P ( x - / ) < ( 250 - 268 / 15)

P ( z < -18 / 15 )

P ( z < -1.2)

= 0.1151

Probability = 0.1151

b ) P ( 250 < x < 280 )

P ( 250 - 268 / 15) < ( x -  / ) < ( 280 - 268 / 15)

P ( - 18 / 15 < z < 12 / 15 )

P (-1.2 < z < 0.8 )

P ( z < 0.8 ) - P ( z < -1.2)

Using z table250

= 0.7881 - 0.1151

= 0.6730

Probability = 0.6730

c ) P(Z < z) = 90%

P(Z < z) = 0.90

P(Z < 1.282) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 15 + 268

x = 287.2