Question

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. Find the probability of a pregnancy lasting 307 days or longer. (ROUND TO 4 DECIMAL PLACES)

b. If the length of pregnancy is in the lowest 44​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

Solution :

Given that ,

mean = = 268 days

standard deviation = = 15 days

a) P(x 307) = 1 - P(x   307)

= 1 - P[(x - ) / (307 - 268 ) / 15]

= 1 -  P(z 2.60)   

  Using z table,

= 1 - 0.9953

= 0.0047

b) Using standard normal table,

P(Z < z) = 44%

= P(Z < z ) = 0.44

= P(Z < -0.15 ) = 0.44  

z = -0.15

Using z-score formula,

x = z * +

x = -0.15 * 15 + 268

x = 265.75

The length that separates premature babies from those who are not premature IS 266 days